While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you,in front of you,or back in your hands if the car is

(a) traveling at a constant speed,
(b) increasing in speed,and
(c) decreasing in speed

Answers

Answer 1

Answer:

a) In the hand

b) behind the hand

c) in front of the hand

Explanation:

a)

When traveling at a constant speed when an egg is tossed directly upwards it lands back in our hand.  It is so because the instant when the egg is released from the hand the egg has the same velocity as the velocity of the person sitting in the moving car, and hence by the virtue of inertia of motion it land lands back in our hand.

b)

When we toss an egg while the car is accelerating then the egg lands behind the hand because the hand along with the car is accelerating ahead. The instant when the egg is released from the hand the car is accelerating then the car has a lower velocity which the egg acquires in the air, while the  egg is in the air the car further gains the velocity and hence the egg lands behind the hand.

C)

When the speed of the car is decreasing then the egg lands in front of the hand because the moment when the egg was tossed into the air it acquired the instantaneous velocity of the car from that moment and then while it was in the air it had the same velocity as initial while the car further decelerated.


Related Questions

A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.320 m/s2. Suppose it moves as a particle under constant acceleration for 4.50 s. Find (a) its position and (b) its velocity at the end of this time interval. Next, assume it moves as a particle in simple harmonic motion for 4.50 s and x 5 0 is its equilibrium position. Find (c) its position and (d) its velocity at the end of this time interval.

Answers

The position and velocity of the particle in equilibrium position at the given parameters are;

A) x = -2.34 m

B) v_x = -1.3 m/s

C) x(4.5) = -0.076 m

D) v = 0.314 m/s

We are given;

Initial distance; x_i = 0.27 m

Initial velocity; v_xi = 0.14 m/s

Acceleration; a_x = - 0.32 m/s²

Time; t = 4.5 s

A) Formula for the particle's position as a function of time under constant acceleration is;

x = x_i + v_xi•t + ½a_x•t²

x = 0.27 + (0.14 × 4.5) + ½(-0.32 × 4.5²)

x = -2.34 m

B) Formula for it's velocity at the end of the time interval is;

v_x = v_xi + a_x•t

v_x = 0.14 + (-0.32 × 4.5)

v_x = -1.3 m/s

C) Formula for position in simple harmonic motion is;

x(t) = A cos(ωt + ϕ)

We know that acceleration is;

a = -ω²x

Thus;

-0.32 = -ω²(0.27)

ω = √(0.32/0.27)

ω = 1.089 rad/s

Now, velocity is the derivative of x(t). Thus;

v(t) = x'(t) = -Aω sin (ωt + ϕ)

At t = 0, we have;

0.14 = -A(1.089) × sin ϕ  - - -(1)

Also, at t = 0,

0.27 = A cos ϕ  - - - (2)

Divide equation 1 by equation 2 to get;

0.14/0.27 = -1.089 tan ϕ

ϕ = tan^(-1) (0.14/(0.27 × -1.089))

ϕ = -25.46°

Thus, putting -25.46° for ϕ in eq 2 gives;

0.27 = A cos (-25.46)

0.27 = A × 0.90183

A = 0.27/0.90183

A = 0.2994

Thus,

x(4.5) = 0.2994 cos((1.089 × 4.5) + (-25.46))

x(4.5) = -0.076 m

D) v = -0.2994 × 1.089 × sin 254.6

v = 0.314 m/s

Read more at; https://brainly.com/question/15089829

Final answer:

The position of the particle after 4.50 s of constant acceleration is 207.360 m, and its velocity is 91.580 m/s. In simple harmonic motion, the position of the particle at the end of 4.50 s is 0 m, and its velocity is -1.080 m/s.

Explanation:

(a) Position:

To find the position of the particle after 4.50 s, we can use the equation x = x0 + v0t + 0.5at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time interval.

Plugging in the given values:

x = 0.270 m + (0.140 m/s)(4.50 s) + 0.5(20.320 m/s^2)(4.50 s)^2

x = 0.270 m + 0.630 m + 206.460 m

x = 207.360 m

Therefore, the position of the particle at the end of 4.50 s is 207.360 m.

(b) Velocity:

To find the velocity of the particle at the end of 4.50 s, we can use the equation v = v0 + at, where v is the final velocity.

Plugging in the given values:

v = 0.140 m/s + (20.320 m/s^2)(4.50 s)

v = 0.140 m/s + 91.440 m/s

v = 91.580 m/s

Therefore, the velocity of the particle at the end of 4.50 s is 91.580 m/s.

(c) Position:

Since the particle is in simple harmonic motion, its position can be described by the equation x = x0 + A* sin(ωt + φ), where x0 is the equilibrium position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

Plugging in the given values:

x = 0 + 0.270*sin(4.00 rad/s * 4.50 s + 0)

x ≈ 0.270*sin(18.000 rad)

x ≈ 0.270*sin(π rad)

x ≈ 0 m

Therefore, the position of the particle at the end of 4.50 s in simple harmonic motion is 0 m.

(d) Velocity:

The velocity of the particle in simple harmonic motion can be described by the equation v = A*ω*cos(ωt + φ).

Plugging in the given values:

v = 0.270*4.00*cos(4.00 rad/s * 4.50 s + 0)

v ≈ 1.080*cos(18.000 rad)

v ≈ 1.080*cos(π rad)

v ≈ -1.080 m/s

Therefore, the velocity of the particle at the end of 4.50 s in simple harmonic motion is -1.080 m/s.

Learn more about Particle motion here:

https://brainly.com/question/23148906

#SPJ12

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera from the railing of the basket when it is 18 mm above the ground. If the balloon continues to rise at the same speed, how high is the railing when the camera hits the ground?

Answers

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

[tex]h = v_0 t -\frac{1}{2} gt^2[/tex]

[tex]-18 = 15*t + \frac{1}{2} 9.8*t^2[/tex]

[tex]t = 3.98s[/tex]

Then the total distance traveled would be

[tex]h = h_0 +v_0t[/tex]

[tex]h = 18+15*3.98[/tex]

[tex]h = 77.7m[/tex]

Therefore the railing will be at a height of 77.7m when it has touched the ground

The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

Calculation o fthe height:

Since we know that

[tex]h = vt - 1/2gt^2\\\\-18 = 15*1 + 1/2*9.8*t^2[/tex]

t = 3.98s

Now the total distance should be

[tex]= 18 + 15*3.98[/tex]

= 77.7 m

hence, The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

Learn more about height here: https://brainly.com/question/24923362

A spherical surface surrounds a point charge q. Describe what happens to the total flux through the surface if the following happens:

(a) The charge is tripled.
*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(b) The volume of the sphere is doubled.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(c) The surface is changed to a cube.
*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(d) The charged is moved to another location inside the surface.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

(e) The charge is moved outside the surface.

*The flux is tripled.
*The flux decreases by 1/3.
*The flux remains constant.
*The flux goes to zero.

Answers

The charge is tripled, flux is tripled.The volume of the sphere is doubled, flux remains constant.The surface is changed to a cube, flux remains constant.The charge is moved to another location inside the surface, flux remains constant.The charge is moved outside the surface, flux goes to zero.

Gauss' law states that the total charge contained within a closed surface immediately proportionately affects the electric flux through that surface. The flux across the spherical surface will quadruple if the charge is tripled.

The enclosed charge determines the electric flux through a closed surface, not the size or shape of the surface. The quantity of charge confined does not change when the sphere's volume is doubled, hence the flux does not change.

Similar to component (b), the flux is unaffected by changing the surface's shape as long as the enclosed charge stays constant.

The total charge contained by the surface is the only factor that affects the electric flux across a closed surface.

Thus, since there is no longer any charge contained by the closed surface if the charge is transported outside of it, there is no longer any electric flux through the surface.

For more details regarding electric flux, visit:

https://brainly.com/question/30409677

#SPJ12

Final answer:

The total flux through a spherical surface surrounding a point charge depends on the charge, while the flux through the surface remains constant when the volume of the sphere or the shape of the surface is changed. If the charge is moved to another location inside the surface, the total flux remains constant, but if the charge is moved outside the surface, the total flux goes to zero.

Explanation:

(a) When the charge is tripled, the total flux through the surface also triples. This is because the total flux is directly proportional to the charge enclosed by the surface.

(b) When the volume of the sphere is doubled, the total flux through the surface remains constant. This is because the total flux is independent of the volume of the surface.

(c) When the surface is changed to a cube, the total flux through the surface remains constant. This is because the total flux is independent of the shape of the surface.

(d) When the charge is moved to another location inside the surface, the total flux through the surface remains constant. This is because the total flux is independent of the position of the charge within the surface.

(e) When the charge is moved outside the surface, the total flux through the surface goes to zero. This is because the total flux is zero when there are no charges enclosed by the surface.

Learn more about Electric Field and Flux here:

https://brainly.com/question/33167286

#SPJ3

At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.41 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

Answers

To develop this problem we will apply the linear motion kinematic equations. For this purpose we will define the change in speed as the product between acceleration and time.

[tex]v_f-v_i = at[/tex]

The relation between initial velocity final velocity and time is

[tex]v_f = v_i+at[/tex]

The acceleration is due to the acceleration due to gravity, then we have

[tex]v_f = v_i-gt[/tex]

At the maximum height the final velocity is zero. Then we have that

[tex]0 = v_i-gt[/tex]

[tex]t = \frac{v_i}{g}[/tex]

The time the player must wait before touching he ball is

[tex]t = \frac{4.41}{9.8}[/tex]

[tex]t = 0.45s[/tex]

An object's velocity in m/s is given by the equation vx(t) = 10t-2. If it starts at x=0 at t=0, what is its position after 5 seconds?

Answers

Answer: d(5) = 115m

Position after 5 seconds is 115m

Explanation:

Given;

Initial position d(0) = 0

Time = 5 sec

Velocity function vx(t) = 10t - 2

To determine its position after 5 sec we need to calculate the position function.

d(t) = integral of vx(t)

d(t) = ∫10t - 2

d(t) = (10/2)t^2 - 2t + c

d(t) = 5t^2 - 2t + d(0)

c = d(0) = 0

d(t) = 5t^2 - 2t

So, at time t = 5

d(5) = 5(5^2) -2(5) = 125 - 10

d(5) = 115m

A 19.2 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.1 . The angle between the horizontal and
the ladder is θ . The person wants to climb
up the ladder a distance of 0.49 m along the
ladder from the ladder’s foot.
What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 0.49 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s^2
Answer in units of ◦

Answers

Answer:

63°

Explanation:

Draw a free body diagram of the ladder.  There are 4 forces:

Normal force N pushing up at the base of the ladder.

Friction force Nμ pushing right at the base of the ladder.

Weight force mg pushing down a distance x up the ladder.

Reaction force R pushing left at the top of the ladder.

Sum of forces in the x direction:

∑Fₓ = ma

Nμ − R = 0

R = Nμ

Sum of forces in the y direction:

∑Fᵧ = ma

N − mg = 0

N = mg

Sum of moments about the base of the ladder:

∑τ = Iα

R (L sin θ) − mg (x cos θ) = 0

R (L sin θ) = mg (x cos θ)

Substituting:

Nμ (L sin θ) = mg (x cos θ)

mgμ (L sin θ) = mg (x cos θ)

μ (L sin θ) = x cos θ

tan θ = x / (μL)

θ = atan(x / (μL))

Given x = 0.49 m, μ = 0.1, and L = 2.5 m:

θ = atan(0.49 m / (0.1 × 2.5 m))

θ ≈ 63°

To solve the problem we will first calculate the reaction and the normal force.

The angle of the ladder should be 63°.

Given to us

Distance the person wants to travel, x = 0.49 m,the coefficient of static friction, μ = 0.1, Length of the ladder, L = 2.5 m:

Free Body DiagramThere are 4 forcesReaction force R pushing left at the top of the ladder,Normal force N pushes up the ladder at the base of the ladder,Friction force Nμ pushing right at the base of the ladder,Weight force of the person pushing down = mg,

Sum of Vertical vector forces,

[tex]\sum F_y = 0\\N - mg = 0\\N = mg[/tex]

Sum of Horizontal vectors forces,

[tex]\sum F_x = 0\\N\mu - R = 0\\R = N\mu[/tex]

Sum of moments at the base of the ladder

[tex]R (L\ sin\theta) - mg (x\ cos\theta) = 0\\R (L\ sin\theta) = mg (x\ cos\theta)[/tex]

Substituting the values of R and N,

[tex]N\mu (L\ sin \theta) = mg (x\ cos \theta)\\mg\mu (L\ sin \theta) = mg (x\ cos\theta)\\\mu (L\ sin \theta) = x\ cos \theta\\tan \theta = \dfrac{x}{\mu L}\\\theta = tan^{-1}( \dfrac{x}{\mu L})[/tex]

Substituting the values,

[tex]\theta = tan^{-1}(\dfrac{0.49\ m}{0.1\times 2.5 m})\\\\\theta = tan^{-1} (0.96)\\\\\theta = 62.969^o \approx 63^o[/tex]

Hence, the angle of the ladder should be 63°.

Learn more about Friction:

https://brainly.com/question/13357196

If an evil genius decided to free the Earth from the Sun by charging both (with an equal charge) to generate an electrical force equal to the gravitational force between them, how much charge would be needed on each?

Answers

Answer:

[tex]2.96866\times 10^{17}\ C[/tex]

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

r = Distance between the objects and particles

[tex]q_1=q_2[/tex] = Charges

M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]

Here, the Electric force will balance the gravitational force

[tex]\dfrac{GMm_2}{r^2}=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{GMm}{k}}\\\Rightarrow q=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}\times 1.989\times 10^{30}}{8.99\times 10^{9}}}\\\Rightarrow q=2.96866\times 10^{17}\ C[/tex]

Charge on each particle will be [tex]2.96866\times 10^{17}\ C[/tex]

You were driving a car with velocity <19, 0, 23> m/s. You quickly turned and braked, and your velocity became <14, 0, 26> m/s. The mass of the car was 1300 kg. (a) What was the (vector) change in momentum during this maneuver? Pay attention to signs. < -6500 , 0 , 3900 > kg·m/s (b) What was the (vector) impulse applied to the car?

Answers

Answer:

a.<-6500,0,3900>kgm/s

b.<-6500,0,3900>kgm/s

Explanation:

We are given that

Initial velocity of car,u=[tex]<19,0,23>[/tex]m/s

Final velocity of car=[tex]v=<14,0,26>m/s[/tex]

Mass of the car=m=1300 kg

a.We have to find the change in momentum during this manuver.

Change in momentum=[tex]\Delta P=m(v-u)[/tex]

Using the formula

[tex]\Delta P=1300(<14,0,26>-<19,0,23>)=1300(<-5,0,3>)=<-6500,0,3900>kgm/s[/tex]

Hence, the change in momentum during this maneuver=<-6500,0,3900>kgm/s

b.Impulse =Change in momentum of car

Impulse applied to the car=<-6500,0,3900>kgm/s

A particle is moving along a straight line such that its' acceleration is defined as
a(v) = (-2v) m/s^2 where v is in meters per second.
If v = 20 m/s when s= 0 and t=0 find:

1. The particles position as a function of time
2. The particles velocity as a function of time
3. The particles acceleration as a function of time

Answers

Final answer:

To find the particle's dynamics, we need to integrate the given acceleration function in terms of velocity and apply the initial conditions. The velocity function v(t) is √(400 - 2t), the position function s(t) would result from further integration, and the acceleration a(t) is -2√(400 - 2t).

Explanation:

A particle is moving in a straight line with its acceleration a(v) given as a(v) = (-2v) m/s², where v is the velocity in meters per second. To find the particle's position, velocity, and acceleration as functions of time, we'll need to integrate the acceleration function.

Finding the velocity as a function of time:

We have the acceleration in terms of velocity; thus, we can write:

'(v) = a(v) = -2v

(v) = -∫ 2v dv = -v² + C

When t=0, v = 20 m/s:

C = 20² = 400

v(t) = √(400 - 2t)

Finding the particle's position as a function of time:

s(t) = ∫ v(t) dt

s(t) = ∫ √(400 - 2t) dt

The integration will give us the position, s(t), in terms of t, which needs to be integrated carefully using appropriate techniques such as substitution.

Finding the acceleration as a function of time:

a(t) can be found by substituting the expression for v(t) into a(v), which gives us a(t) = -2√(400 - 2t).

Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.
What is the magnitude of the electrostatic force acting between them?

Answers

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

[tex]F=\frac{kq^{2} }{r^{2} }[/tex]

Substitute the given values we get

[tex]F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N[/tex]

A 5.60 g bullet moving at 501.8 m/s penetrates a tree trunk to a depth of 4.59 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet. Answer in units of N

Answers

Answer:

Explanation:

Given

mass of bullet [tex]m=5.6\ gm[/tex]

velocity of bullet [tex]v=501.8\ m/s[/tex]

Depth of penetration [tex]d=4.59\ cm[/tex]

According to the work energy theorem work done by all the force will be equal to change in kinetic energy of Particle

Suppose F is the force which is opposing the bullet motion

change in kinetic Energy [tex]\Delta K=\frac{1}{2}mv^2-0[/tex]

[tex]\Delta K=\frac{1}{2}mv^2=\frac{1}{2}\times 5.6\times 10^{-3}\times (501.8)^2[/tex]

[tex]\Delta K=705.049\ J[/tex]

[tex]\Delta K=F\cdot d[/tex]

[tex]F=\frac{\Delta K}{d}[/tex]

[tex]F=\frac{705.049}{4.59\times 10^{-2}}[/tex]

[tex]F=15,360.54\ N[/tex]

[tex]F=1.536\times 10^4\ N[/tex]                      

An 80-cm-long, 1.0-mm-diameter steel guitar string must betightened to a tension of 2000 {\rm N} by turning the tuningscrews.By how much is the string stretched? (In 2 sig figs and cm)

Answers

Answer:

string stretched is 1.02 cm

Explanation:

given data

length = 80-cm

diameter = 1.0-mm

tension = 2000 N

solution

we get here string stretched that will be as and here  

we know that young modulus for steel = 200 × [tex]10^{9}[/tex]

so here stress will be

stress = y × strain  .............1

that is express as

[tex]\frac{force}{area} = \frac{Y \Delta L}{L}[/tex]

ΔL = [tex]\frac{0.80*2000}{\pi * 0.0005^2*200*10^9}[/tex]

ΔL = 0.0102 m

ΔL = 1.02 cm

so string stretched is 1.02 cm

Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The surface temperature is about 37 K.

(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?

(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).

(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)

Answers

Answer:

a) The number density is 3.623 × 10⁻³ [tex]\frac{mol}{m^{3} }[/tex]

The mass of the atmosphere is 1.3 × 10²²Kg

b) The pressure is 10⁻²⁰ Millimeter of mercury

c) The mass mixing ratio is 0.0107

The partial pressure of ethane is 0.01114 Pa

Yes it is condensable because it boiling point is -88.5 C  which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere  is 37 K.

Explanation:

The explanation is on the first and second uploaded image

A cylindrical shell of length 220 m and radius 4 cm carries a uniform surface charge density of σ = 14 nC/m^2.
1. What is the total charge on the shell?

Answers

Answer:

Explanation:

Given

Length of shell [tex]L=220\ m[/tex]

radius of cylindrical shell [tex]r=4\ cm[/tex]

surface charge density [tex]\sigma =14\ nC/m^2[/tex]

Total charge on the shell [tex]Q=surface\ area\times surface\ charge\ density[/tex]

surface area [tex]A=2\pi r\cdot L=2\pi \cdot 0.04\cdot 220=55.299\ m^2[/tex]

[tex]Q=\sigma \cdor A[/tex]

[tex]Q=14\times 10^{-9}\times 55.299[/tex]

[tex]Q=7.741\times 10^{-7}\ C[/tex]

A circuit consists of a 12.0-V battery connected to three resistors (44ohm , 17ohm and 110ohm ) in series.a. Find the current that flows through the battery.I=__mAb.Find the potential difference across the 44ohm resistor.V1=__Vc.Find the potential difference across the 17ohm resistor.V2=__Vd.Find the potential difference across the 110ohm resistor.V3=__V

Answers

Answer:

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Explanation:

Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

From ohm's law

Voltage (V) = current (I) × resistance (R)

I = V/R = 12/171 = 0.0702A = 0.0702×1000mA = 70.2mA

V1 = I×R1 = 0.0702×44 = 3.09V

V2 = I×R2 = 0.0702×17 = 1.19V

V3 = I×R3 = 0.0702×110 = 7.72V

Answer:

a) I = 0.0702A = 70.2mA

b) V1 = 3.09V

c) V2 = 1.19V

d) V3 = 7.72V

Explanation:

Given that the circuit consist of three series resistors.

For resistors arranged in series, the total resistance R can be given as:

R = R1 + R2 + R3

R1 = 44 ohms

R2 = 17 ohms

R3 = 110 ohms.

R = 44 + 17 + 110 = 171 ohms

V = 12 V

a) The current of a circuit is given by;

Potential difference V = current × total resistance

V = IR

Making I the subject of formula,

I = V/R

I = 12/171 = 0.0702A

I = 7.02mA

b) the potential difference across any resistors is given by:

V = IR

Since the arrangement is parallel, the same current flows through each of the resistors.

V1 = IR1

V1 = 0.0702 × 44

V1 = 3.09V

c) applying the same rule as b above:

V2 = IR2

V2 = 0.0702 × 17

V2 = 1.19V

d) applying the same rule as b above.

V3 = IR3

V3 = 0.0702 × 110

V3 = 7.72V

If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many imageswould you see in the viewer?

Answers

The correct answer is  (a.) thousands of small images. Each hole in the aluminum foil acts as a tiny pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one forms a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.

If you punched thousands of holes in the aluminum foil of the scope (so there were more “holes” than “foil”), the resulting images in the viewer would be thousands of small images. Each hole acts as a pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one would form a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.

complete question;

If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images would you see in the viewer?

Choices are:

a. thousands of small images

b. a few bright images

c. one large blurry image

d. no images at all, the light waves would cancel

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant force acts on the particle, with Fx=-7N and Fy= +5N.
What is the magnitude of the momentum of the particle at the end of this 0.13-second interval?

Answers

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

[tex]|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s[/tex]

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 18 m/s at an angle 42 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 14 m/s when it reaches a maximum height of 10 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
6) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

Answers

Answer:

1) 13.377 m/s

2) 12.044 m/s

3) 8.893 m

4) 32.85 m

5) 19.05 m/s

6) 3.25 m

Explanation:

1)

V_o,x = V_o * cos (Q)

V_o,x = 18 * cos (42)

V_o,x = 13.377 m/s

2)

V_o,y = V_o * sin (Q)

V_o,y = 18 * sin (42)

V_o,y = 12.044 m/s

3)

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 = 12.044 - 9.81t

Solve above equation for t:

t = 1.228 s

Compute S_y @t = 1.228 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 1.5 + 12.044*1.228 - 4.905*1.228^2

S_y = 8.893 m

4)

Time taken for the ball to complete path:

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 1.5

V_o,y*t + 0.5*a*t^2 = 0

12.044*t - 4.905*t^2 = 0

t = 0, t = 2.455 s

Total distance traveled in horizontal direction S_x @ t = 2.455 s

S_x = S_o,x + V_o,x*t

S_x = 0 + 13.377*2.455 = 32.85 m

5)

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 10

10 = 1.5 + V_o,y*t -4.905*t^2   .... Eq 1

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 =  V_o,y - 9.81t  .... Eq2

Solve Eq 1 and Eq 2 simultaneously

V_o,y = 9.81*t

10 = 1.5 + 9.81*t^2 -4.905*t^2

8.5 = 4.905*t^2

t = 1.316 s

V_o,y = 12.914 m/s

Compute Velocity

V = sqrt (V_o,x^2 + V_o,y^2)

V = sqrt (14^2 + 12.914^2)

V = 19.05 m/s

6)

Total distance traveled in horizontal direction between players is 32.85m

S_x = S_o,x + V_o,x*t

S_x = 0 + 14*t = 32.85 m

t = 2.3464 s

Compute Sy @ t = 2.3464 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 10  - 4.905*(1.1732)^2

S_y = 3.25 m

A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.00 m/s?

Answers

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

[tex]v = r\omega \rightarrow \omega = \frac{v}{r}[/tex]

Here,

v = Lineal velocity

[tex]\omega[/tex]= Angular velocity

r = Radius

Our values are

[tex]v = 6/ms[/tex]

[tex]r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m[/tex]

Replacing to find the angular velocity we have,

[tex]\omega = \frac{6m/s}{0.06m}[/tex]

[tex]\omega = 100rad/s[/tex]

Convert the units to RPM we have that

[tex]\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})[/tex]

[tex]\omega = 955.41rpm[/tex]

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

Final answer:

The angular speed of the scooter's wheels when the scooter is moving at 6.00 m/s is 100 rad/sec. This value is calculated using the formula for relating linear velocity, radius, and angular speed, and converting the wheel's diameter to a radius in meters.

Explanation:

To find the angular speed of the scooter's wheels, we need to use the equation that relates linear velocity (v), radius (r), and angular speed (w). This equation is v = r*w where v is the linear speed, r is the radius, and w is the angular speed which we are trying to find.

First, radius r needs to be calculated using the provided diameter as r = diameter / 2 = 120 mm / 2 = 60 mm. Since the linear speed is provided in m/s, we need to convert the radius from mm to m. So, r = 60 mm = 0.06 m.

Then, we can substitute the known values into the equation. 6 m/s = 0.06 m * w, and solve for w: w = (6 m/s) / 0.06 m = 100 rad/s. Therefore, the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 100 rad/sec.

Learn more about Angular Speed here:

https://brainly.com/question/32114693

#SPJ12

Two parallel plates have equal but opposite charges on their surface. The plates are separated by a finite distance.

A fast moving proton enters the space between the two plates through a tiny hole in the left plate A. The electric potential energy of the proton increases as it moves toward plate B.

(a) How is the speed of the proton affected as it moves from plate A to plate B?
increases
decreases
stays the same

(b) Which plate is at a higher potential?
plate A
plate B

(c) What can you conclude about the charges on the two plates?
Plate A is positive and plate B is negative.
Plate A is negative and plate B is positive.

(d) What will be the pattern of the equipotential lines in the space between the two plates? (Select all that apply.)
Parallel lines more closely packed near plate B.
Parallel lines perpendicular to the two plates.
Parallel lines parallel to the two plates.
Parallel lines more closely packed near plate A.
Parallel lines equally spaced.

Answers

Answer:

(a) The speed of proton decreases as it moves from A to B.

(b) Plate B is at a higher potential.

(c) Plate B is positive, plate A is negative.

(d) Parallel lines parallel to the two plates.

    Parallel lines equally spaced.

Explanation:

The electric potential energy is given by the following formula:

[tex]U = \frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}[/tex]

Alternatively, potential energy in a uniform electric field is

[tex]U = qEr[/tex]

where 'r' is the distance from negative to positive plates. This definition is analogues to that of gravitational potential energy, U = mgh.

If the positively charged proton is gaining potential energy as it gets closer to plate B, then plate B is charged positively.

(a) According to this information, the speed of proton decreases as it moves from A to B. This is similar to the speed of an object which is gaining potential energy by moving higher.

(b) By the same gravitational analogy, plate B is at a higher potential.

(c) As explained before, Plate A is negative and Plate B is positive.

(d) The equipotential lines are parallel to electric field lines which are perpendicular to the plates. So, the equipotential lines are parallel to the plates. Since the electric field between the plates is uniform, then the equipotential lines are equally seperated.

A 50 kg child runs off a dock at 2.0 ms (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the rowboat move away from the dock?

Answers

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

Final answer:

The child running off a dock and landing in a rowboat scenario involves applying the conservation of momentum principle to find the boat's final velocity. The speed of the rowboat is 0.67 m/s.

Explanation:

Given:

Child mass (m1) = 50 kg

Child velocity (v1) = 2.0 m/s

Boat mass (m2) = 150 kg

Let the final velocity of the boat be v2

Using the conservation of momentum:

m1v1 = (m1 + m2)v2

Substitute the values to find v2: 50 kg * 2.0 m/s = (50 kg + 150 kg) * v2

Solving for v2, we get v2 = 0.67 m/s

Learn more about Conservation of Momentum here:

https://brainly.com/question/35273367

#SPJ11

How much work is done by the motor in a CD player to make a CD spin, starting from rest? The CD has a diameter of 12.70 cm and a mass of 16.30 g. The laser scans at a constant tangential velocity of 1.150 m/s. Assume that the music is first detected at a radius of 20.90 mm from the center of the disk. Ignore the small circular hole at the CD's center.

Answers

Final answer:

The work done by the motor in a CD player to make the CD spin can be calculated using the formula: Work = Torque * Angle. The final angular velocity can be calculated using the formula: v = r * ω. The moment of inertia can be calculated using the formula: Moment of Inertia = (1/2) * Mass * Radius^2.

Explanation:

The work done by the motor in a CD player to make the CD spin can be calculated using the formula:

Work = Torque * Angle

In this case, since the CD starts from rest, the initial angular velocity is zero. The final angular velocity can be calculated using the formula:

v = r * ω

where r is the radius of the CD and ω is the angular velocity.

The torque can be calculated using the formula:

Torque = Moment of Inertia * Angular Acceleration

Since we have the mass and diameter of the CD, the moment of inertia can be calculated using the formula:

Moment of Inertia = (1/2) * Mass * Radius^2

Once we have the torque and the angle, we can calculate the work done by the motor.

How much time elapses from when the first bit starts to be created until the conversion back to analog begins

Answers

The question is not complete. Kindly find the complete question below:

Host A converts analog to digital at a = 58 Kbps  Link transmission rate R = 1.9 Mbps  Host A groups data into packets of length L = 112 bytes  Distance to travel d = 931.9 km  Propagation speed s = 2.5 x 108 m/s  Host A sends each packet to Host B as soon as it gathers a whole packet.  Host B converts back from digital to analog as soon as it receives a whole packet.  How much time elapses from when the first bit starts to be created until the conversion back to analog begins? Give answer in milliseconds (ms) to two decimal places, normal rounding, without units (e.g. 1.5623 ms would be entered as "1.56" without the quotes)

Answer / Explanation

The answer is 19.65

The pressure reading from a barometer is 742 mm Hg. Express this reading in kilopascals, kPa. (Use 760 mm Hg = 1.013 x 105 Pa)

Answers

Answer:

98.9 kPa

Explanation:

given,

Pressure reading of barometer = 742 mm Hg

we know,

760 mm Hg = 1.013 x 10⁵ Pa

[tex]1\ mm\ Hg = \dfrac{1.013\times 10^5}{760}[/tex]

[tex]742\ mm\ Hg = \dfrac{1.013\times 10^5}{760}\times 742[/tex]

                           = 0.989 x 10⁵ Pa

                          = 98.9 x 10³ Pa

                          = 98.9 kPa

the reading of the barometer is equal to  98.9 kPa

The pressure reading from the barometer expressed in kilopascal is 98.9kPa.

Given that;

Pressure reading from the barometer; [tex]P = 742mmHg[/tex]

Pressure reading from the barometer in kilopascals; [tex]x = \ ?[/tex]

First we convert the units from Millimeter of Mercury (mmHg) to Pascal (Pa)

We are to use;

[tex]760 mm Hg = 1.013 * 10^5 Pa\\\\\frac{760mmHg}{760} = \frac{1.013 * 10^5 Pa}{760} \\\\1mmHg = 1.33289 * 10^2 Pa[/tex]

So,

Pressure reading is pascal

[tex]P = 742 * [1.33289*10^2Pa]\\\\P = 98900.438Pa[/tex]

Next we convert to kilopascal

We know that; [tex]1\ Pascal = 0.001\ kilopascal[/tex]

so

[tex]P = 98900.438 * 0.001 kPa\\\\P = 98.9kPa[/tex]

Therefore, the pressure reading from the barometer expressed in kilopascal is 98.9kPa.

Learn more: https://brainly.com/question/9844551

A boat is able to move throught still water at 20m/s. It makes a round trip to a town 3.0km upstrea. If the river flows at 5m/s, the time required for this round trip is

Answers

Answer:

t=320s

Explanation:

Given Data

Boat speed=20 m/s

River flows=5 m/s

Total trip of distance d=3.0km = 3000m

To find

Total time taken

Solution

As

[tex]Velocity=distance/time\\time=distance/velocity\\[/tex]

Here we have two conditions

First when boat moves upward and the river pushing back.then velocity is given as

velocity=20m/s-5m/s

velocity=15 m/s

Time for that velocity

[tex]t_{1} =distance/velocity\\t_{1}=\frac{3000m}{15m/s}\\ t_{1}=200s[/tex]

Now for second condition when river flows and boat speed on same direction

velocity=20m/s+5m/s

velocity=25 m/s

Time taken for that velocity

[tex]t_{2}=distance/velocity\\t_{2}=\frac{3000m}{25m/s}\\ t_{2}=120m/s[/tex]

Now the total time

[tex]t=t_{1}+t_{2}\\t=(200+120)s\\t=320s[/tex]

A mass m at the end of a spring oscillates with a frequency of 0.83 Hz.When an additional 730 gmass is added to m, the frequency is 0.65 Hz.What is the value of m?

Answers

Answer:

m will be equal to 1158.73 gram

Explanation:

We have given mass m when frequency is 0.83 Hz

So mass [tex]m_1=m[/tex] and frequency [tex]f_!=f[/tex] let spring constant of the spring is KK

Frequency of oscillation of spring is given by [tex]f=\frac{1}{2 \pi }\sqrt{\frac{k}{m}}[/tex]

From above relation we can say that [tex]{\frac{f_1}{f_2}}=\sqrt{\frac{m_2}{m_1}}[/tex]

It is given that when an additional 730 gram is added to m then frequency become 0.65 Hz , [tex]f_2=0.65Hz[/tex]

So [tex]m_2=m+730[/tex]

So  [tex]\frac{0.93}{0.65}=\sqrt{\frac{m+730}{m}}[/tex]

[tex]\frac{m+730}{m}=1.63[/tex]

[tex]0.63m=730[/tex]

m= 1158.73 gram

Final answer:

To find the value of m, we can set up a proportion using the formula for frequency and solve for m. The value of m is 0.93 kg.

Explanation:

To solve this problem, we can use the formula for the frequency of an object in simple harmonic motion:

f = 1 / T

Where f is the frequency and T is the period. Let's denote the mass of the object as m, and the original frequency as f1. When the additional mass is added, the frequency becomes f2. We can set up a proportion to solve for the value of m:

f1 / f2 = m / (m + 0.73)

Solving for m, we have:

m = (f1 / f2) * 0.73

Substituting the values f1 = 0.83 Hz and f2 = 0.65 Hz, we can find the value of m:

m = (0.83 / 0.65) * 0.73 = 0.93 kg

Therefore, the value of m is 0.93 kg.

Three point charges are arranged in a line. Charge q 3 = +5 nC is at the origin. Charge q 2 = -3 nC is located at x- +4.00 cm. Charge q 1 is at x= + 2.00 cm. What is q 1, in magnitude and sign, if the net force on q3 is zero?

Answers

Answer:

q1= +0.75 nC

Explanation:

As the electrostatic force is linear, we can apply the superposition principle to calculate the total force on q₃ due to q₂ and q1, according to Coulomb's Law, as follows:

F₃₂ = k*q₃*q₂/r₃₂² = 9*10⁹ N*m²/C²*+5nC*(-3 nC) / (0.04m)² = -84.4*10⁻⁶ N

F₃₁ = k*q₃*q₁ / r₃₁² = 9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

The total force on q₃ is just the sum of F₃₂ and F₃₁, which must add to 0, as follows:

F₃ = F₃₂ + F₃₁ = 0

⇒ -84.4* 10⁻⁶ N = -9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

Solving for q₁, we get:

q₁ = (84.4 / 11.25)*10⁻¹⁰ C = +0.75 nC

q₁ must be positive, in order to counteract the attractive force on q₃ due to q₂.

A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the sphere has a surplus charge of 4 × 10 4 units. What is the charge on the half of the rod farther from the sphere? 0 − 1 × 10 8 1 × 10 8 − 4 × 10 4 4 × 10 4

Answers

Final answer:

The charge on the half of the rod farther from the sphere brought near to it, which had initially a neutral charge, is -4×104 units. This is due to a process known as charging by induction.

Explanation:

The charge on the half of the rod farther from the sphere is -4×104 units. This occurs due to a process called charging by induction. Basically, when a charged object is brought near to an initially neutral conductor, it polarizes the conductor. Negative charges are attracted towards the charged sphere, leaving the far side of the rod positively charged.

However, the problem statement tells us that there is a surplus of charge on the closer half of the rod, hence the further half must have a deficiency of charge by the same amount, resulting in -4×104 units of charge. Remember, in a neutral object, the total charge is zero. Thus, if we develop a surplus (+4×104) on one side, we must have an equal amount of deficit (-4×104) on the other side to maintain the total charge at zero.

Learn more about Charging by induction here:

https://brainly.com/question/10254645

#SPJ12

A heat engine uses two containers held at different temperatures. One container is at 294 K 294 K , while the other is kept at 552 K 552 K . What is the maximum possible efficiency for this engine?

Answers

Answer:

Explanation:

Given

Lower Temperature [tex]T_L=294 \K[/tex]

Higher Temperature [tex]T_H=522 \K[/tex]

Maximum Possible efficiency is achieved when the engine works as carnot Engine

i.e. [tex]\eta _{max}=1-\frac{T_L}{T_H}[/tex]

[tex]\eta_{max}=1-\frac{294}{522}[/tex]

[tex]\eta _{max}=\frac{228}{522}=0.436[/tex]

[tex]\eta _{max}=43.64\ %[/tex]

Final answer:

The maximum possible efficiency of a heat engine can be determined using the temperatures of the hot and cold reservoirs in the formula Effc = 1 - Tc / Th. Applying this formula to the given temperatures of 294 K and 552 K results in a maximum efficiency of 46.8%.

Explanation:

The maximum possible efficiency of a heat engine (also known as the Carnot efficiency) can be calculated using the temperatures of the heat source (hot reservoir) and the heat sink (cold reservoir). This efficiency can be determined by using the formula Effc = 1 - Tc / Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Both these temperatures should be in Kelvin.

In the given problem, Tc is 294 K and Th is 552 K. Substituting these values into the formula gives the maximum possible efficiency:

Effc = 1 - 294 / 552 = 0.468. Thus, the maximum possible efficiency of this engine is approximately 46.8%.

Learn more about Carnot Efficiency here:

https://brainly.com/question/37665551

#SPJ3

Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 5 in a one-dimensional box 45.7 pm in length.

Answers

Answer: wavelength λ = 2.9Å

Explanation:

Using the particle in a box model. The energy level level increases with n^2

En = (n^2h^2)/ 8mL^2 .....1

For the ground state, n = 1 to level n= 5, the energy level changes from E1 to E5

∆E = (5^2 - 1^2)h^2/8mL^2

but 5^2 - 1^2 = 24.

so,

∆E = 24h^2/8mL^2 .....2

And the wavelength of the radiation can be derived from the equation below:

E = hc/λ

λ = hc/E .......3

Substituting equation 2 to 3

λ = hc/[(24h^2)/ 8mL^2]

λ = 8mcL^2/(24h)

λ = 8mcL^2/24h .....4

Where,

n = energy state

h = Planck's constant = 6.626 × 10^-34 Js

m= mass of electron = 9.1 × 10^-31 kg

L = length = 45.7pm = 45.7×10^-12 m

E = energy

c= speed of light = 3.0 ×10^8 m/s

λ= wavelength

Substituting the values into equation 4 above

λ = [(8×9.1×3×45.7^2)/(24×6.626)] × 10^(-31+8-24+34)

λ = 2868.285390884 × 10^-13 m

λ = 2.9 × 10^-10 m

λ = 2.9Å

Final answer:

The wavelength of the electromagnetic radiation required to excite an electron can be calculated using the formula wavelength = (2 * box length) / n, where n is the energy level. The wavelength of the electromagnetic radiation required is 9.14 pm.

Explanation:

To calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 5 in a one-dimensional box, we can use the formula:

wavelength = (2 * box length) / n

Given that the box length is 45.7 pm and n = 5, we can substitute the values into the formula:

wavelength = (2 * 45.7 pm) / 5

Simplifying the expression gives us:

wavelength = 9.14 pm

Therefore, the wavelength of the electromagnetic radiation required is 9.14 pm.

Learn more about Electromagnetic Radiation here:

https://brainly.com/question/2334706

#SPJ3

Other Questions
How many carbon atoms are there in 5.25 mol of C2H6? After his company closed the retail location he managed, while Tareq was interviewing for other management positions, he began a painting business. Tareq is an example of a(n)___________. Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine randomly selected individuals are given below:108.6 117.4 128.4 120.0 103.7 112.0 98.3 121.5 123.2(a) What is the median of the reported blood pressure values?(b) Suppose the blood pressure of the second individual is 117.9 rather than 117.4 (a small change in a single value). What is the new median of the reported values?(c) What does this say about the sensitivity of the median to rounding or grouping in the data?A. When there is rounding or grouping, the median can be highly sensitive to small change.B. When there is rounding or grouping, the median is only sensitive to large changes.C. When there is rounding or grouping, the median is not sensitive to small changes. 7 movie tickets cost $25.00 Choose the false statement about fission. Group of answer choices A) Fission reactors release harmful gasses into the atmosphere - mainly CO2. B) Fission occurs when an atomic nuclei decay spontaneously or is bombarded by neutrons.. C) Uranium splits into Krypton and Barium D) Fission creates an incredible amount of energy through the conversion of a small amount of mass. Why is it important to understand the context of a source?A. A source's context helps an audience interpret its meaning.OB. Sources are only unbiased if the context is understood.C. Evaluating context is the only way to identify a primary orsecondary source.OD. Without context, a source cannot have a valid point of view. X divided by 4 equals -7. What is x? Which phrase correctly defines the motives of a business? A. satisfying the needs of people B. helping a country's economy grow C. incurring expenses D. earning profits and providing value in the form of goods or services What is the standard deviation (s) of the following set of scores? 10, 15, 12, 18, 19, 16, 12 Student measured the absorbance of the substance X depending on its concentration in solution. She used 0.2 cm cuvette. Her calibration curve is a straight line with the slope 2.3103 M-1. The extinction coefficient of substance X in this solution is: 1. Stephanie Schneider is running down the street at 10 m/s. She sees a double-tall latte onthe road ahead of her and starts sprinting. After 5 seconds have passed, Stephanie isrunning at 15 m/s. Assuming she does not change direction, what was Stephanie'sacceleration? People have different ways of relieving tension. Some of those methods are Which would have a greater impact on the freezing point depression of ice, CrF3 or CaF2? Explain briefly. You randomly select k integers between 1 and 100, inclusive. What is the smallest k that guarantees that at least one pair of the selected integers will sum to 101? Prove your answer. An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volume of water,respectively. For a respiration rate of 40 breaths per minute, how many moles of water per minute are expelled from the body through the lungs? Shay did not turn in her time sheet at work. she says she simply forgot, but her manager has to write up a report anyway. which of the following words could her manager use to show this problem in the most positive way? 1. definite falsehood2. honest mistake3. neglectful behavior4. serious infringement describe the traditions and customs that developed beyond the control of white authority During the year just ended, Shering Distributors, Inc., had pretax earnings from operations of $490,000. In addition, during the year it received $20,000 in income from interest on bonds it help in Zig Manufacturing and received $20,000 in income from dividends on its 5% common stock holding in Tank Industries, Inc. Shering is in the 40% tax bracket and is eligible for a 70% dividend exclusion on its Tank Industries stock.A. Calculate the firm's tax on its operating earnings only.B. Find the tax and after-tax amount attributable to dividend and interest income.b. Interest c. Dividend Before-tax income $0 $0 Less exclusion 0 $0 Use the exclusion 70%Taxable income $0 $0 Tax 40% $0 $0 After-tax amount $0 $0 Compare, contrast, and discuss the after-tax amounts resulting from the interest income and dividend income calculated in parts b. and c. (Select all the choices that apply.) A. The after-tax amount of dividends received, $ 24 comma 165, exceeds the after-tax amount of interest, $ 21 comma 330, due to the 50 % corporate dividend exclusion. B. The after-tax amount of dividends received, $ 21 comma 330, exceeds the after-tax amount of interest, $ 24 comma 165, due to the 50 % corporate dividend exclusion. C. Since the after-tax amount of interest exceeds the after-tax amount of dividends, this increases the attractiveness of stock investments by one corporation in another relative to bond investments. D. Since the after-tax amount of dividends exceeds the after-tax amount of interest, this increases the attractiveness of stock investments by one corporation in another relative to bond investments. The tables below show four sets of data:Set Ax123456789y1098765432Set Bx123456789y34567891011Set Cx123456789y86543.532.522Set Dx123456789y12.52.5345689For which set of data will the scatter plot represent a positive linear association between x and y? Set A Set B Set C Set D Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break the water pouch, the ammonium nitrate dissolves in water and the pack gets cold. The heat of solution for ammonium nitrate is 25.4 kJ/mol.a. Is the dissolution of ammonium nitrate endothermic or exothermic?b. A cold pack contains 135.0 g of water and 50.0 g of ammonium nitrate. What will be the final temperature of the activated cold pack, if the initial temperature is 25.0 degree C? (Assume that the specific heat of the solution is the same as that for water, 4.184 J/g degree C and no heat is lost). Steam Workshop Downloader