can someone help me?!!!!!

Answer:

I believe the answer is B.

Explanation:

Microbes could cause infectious diseases like for Ex: Flu and measles or heart diseases.

Hope my answer has helped you! :)

Microbes are found to be problematic for two reasons that they cause infection and sometime these organisms does not respond to antibiotics.

Explanation:

The microbes can cause infections in humans is a problem. Fungi, helminths and protozoans are the major cause of infectious diseases. Resistance to an antibiotic occurs when an ability to defeat drugs created to kill them are developed in a microbe. When a microbe becomes resistant, the antibiotics cannot fight against them and hence they multiply.  

Antibiotic resistance is an urgent threat to the health of the public. They become a threat, when a microbe become resistant to antibiotic and it will be very difficult to destroy them. The resistance to antibiotics can cause serious issues like disability or even cause death.

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

[tex]sin \theta = \frac{n \lambda}{a}[/tex]

where

[tex]\theta[/tex] is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

[tex]\lambda[/tex] is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because [tex]\theta[/tex] increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) [tex]10.8^{\circ}[/tex]

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

[tex]sin \theta = \frac{(1) \lambda}{a}[/tex]

in this case we have

[tex]\lambda=54 cm = 0.54 m[/tex] is the wavelength

[tex]a=5.7 m[/tex] is the width of the opening

Solving the equation, we find

[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}[/tex]

So the angular spread of the central diffraction maximum is twice this angle:

[tex]\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}[/tex]

(c) [tex]0.26^{\circ}[/tex]

Here we can apply the same formula used before, but this time the wavelength of the signal is

[tex]\lambda=13 mm=0.013 m[/tex]

so the angle corresponding to the first minimum is

[tex]\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}[/tex]

So the angular spread of the central diffraction maximum is twice this angle:

[tex]\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}[/tex]

The change from a 54 cm wavelength to a 13 mm wavelength will decrease the diffraction of signals. The angular spread for a 54 cm wavelength is 0.1154 degrees and for a 13 mm wavelength, it is 0.00279 degrees.

The shift from conventional television signals with a wavelength of about 54 cm to future digital television signals with a wavelength of about 13 mm will decrease the diffraction of the signals into shadow regions of obstacles. This is because diffraction is inversely proportional to wavelength.

The angular spread of the central diffraction maximum can be calculated using diffraction spreading for a single slit given by the equation θ = 1.22 λ / D, where θ is the angle, λ is the wavelength, and D is the slit width. For a wavelength (λ) of 54 cm and a slit width (D) of 5.7 m, the angular spread θ is:

θ = 1.22 * (0.54 m) / 5.7 m = 0.1154 °

For a wavelength of 13 mm, the angular spread θ is:

θ = 1.22 * (0.013 m) / 5.7 m = 0.00279 °

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

[tex]v_f^2 = v_i^2 + 2 a \Delta y[/tex]

where

[tex]v_f[/tex] is the final velocity

[tex]v_i[/tex] is the initial velocity

a is the acceleration

[tex]\Delta y[/tex] is the distance covered

For the particle in free-fall in this problem, we have

[tex]v_i = 0[/tex] (it starts from rest)

[tex]v_f = 7.9 m/s[/tex]

[tex]g=9.8 m/s^2[/tex] (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

[tex]\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m[/tex]

Answer: Last option

They all have the same current.

Explanation:

A connection of three elements in series is represented as follows:

--------[4Ω]-------[8Ω]--------[12Ω]----------

→ I

Note that the three elements share the same current line I .

By definition when the resistors or other electrical components are connected in series then the same current passes through them. Therefore in this case the magnitude of the resistance does not influence the magnitude of the current.

The answer is the last option

Answer:

56.0 m

Explanation:

We know that after 0.510 s, the ball is level with Bob again.  We can use this to find the vertical component of the initial velocity.

y = y₀ + v₀ᵧ t + ½ gt²

h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²

v₀ᵧ = 2.50 m/s

Since the magnitude is 34.1 m/s, we can now find the horizontal component:

v₀² = v₀ₓ² + v₀ᵧ²

(34.1)² = v₀ₓ² + (2.50)²

v₀ₓ = 34.0 m/s

And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:

x = x₀ + v₀ₓ t + ½ at²

126 = 0 + (34.0) t + ½ (0) t²

t = 3.71 s

Finally, we can now find the height of the cliff:

y = y₀ + v₀ᵧ t + ½ gt²

0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²

h = 56.0 m

Bob climbed approximately 4.4 meters.

To determine the height that Bob climbed, we can use the kinematic equation for vertical motion:

Y = Yo + Voy*t -1/2gt^2

where:

Y = final height (unknown)

Yo = initial height (2m)

Voy = initial vertical velocity (unknown)

t = time taken to reach the final height (0.510s)

g = acceleration due to gravity (9.8m/s^2)

We need to find the initial vertical velocity. When the ball reaches the final height, it has the same vertical velocity as it did when it was thrown. Using the equation for vertical velocity:

Vy = Voy - gt

where:

Vy = vertical velocity (0 m/s)

Substituting the known values:

0 = Voy - (9.8)(0.510)

Solving for Voy, we get:

Voy ≈ 5.0 m/s

Now we can calculate the final height:

Y = Yo + Voy*t - 1/2gt^2

Y = 2 + (5.0)(0.510) - 1/2(9.8)(0.510)^2

Y ≈ 2 + 2.55 - 0.126

Y ≈ 4.425m

Therefore, Bob climbed approximately 4.4 meters.

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Answer:

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

Explanation:

For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:

[tex]E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2[/tex] (1)

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

We can also notice that the force on the spring is given by Hook's law:

[tex]F=-kx[/tex]

And since according to Newton's law we have F = ma, this can be rewritten as

[tex]ma=-kx\\a=-\frac{k}{m}x[/tex]

which means that the acceleration is proportional to the displacement.

So by looking again at eq.(1), we can now states that:

- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

- in all the other intermediate situations, both velocity and acceleration are non-zero

So the correct answers are

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

The object oscillating on a spring can have zero acceleration and nonzero velocity, zero velocity and nonzero acceleration, or nonzero velocity and nonzero acceleration at some point during the oscillation. However, it's not possible for it to have both zero velocity and zero acceleration at the same time. These properties are due to the physics of simple harmonic motion.

Let's look at the motion of an object oscillating on a spring, a scenario that involves simple harmonic motion:

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The answer is:

The first option, the force tending to lift Rover is equal to 14.5 N.

To calculate the force that is tending to lift Rover vertically, we need to calculate the vertical component force.

Since we know that the angle between the force and the ground is 29°, we can calculate the vertical component of the force using the following formula:

[tex]F_y=Force*Sin(29\°)[/tex]

We are given that the force is equal to 30.0 N, so, calculating we have:

[tex]F_y=Force*Sin(29\°)[/tex]

[tex]F_y=30N*Sin(29\°)=14.5N[/tex]

Also, we can calculate the horizontal component of the force using the following formula:

[tex]F_x=Force*Cos(29\°)[/tex]

[tex]F_x=30N*Cos(29\°)=26.24N[/tex]

Hence, we have that the correct option is the first option, the force tending to lift Rover is equal to 14.5 N.

Have a nice day!

Answer:

C

Explanation:

Objects in free-fall (also known as projectiles) follow a parabolic curve.  So the answer is C.

Answer:

20 m/s westward

Explanation:

Taking eastward as positive direction, we have:

[tex]v_B = +5 m/s[/tex] is the velocity of Bill with respect to Amy (which is stationary)

[tex]v_C = -15 m/s[/tex] is the velocity of Carlos with respect to Amy

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

[tex]v_B = +5 m/s[/tex]

Therefore, Carlos velocity in Bill's reference frame will be

[tex]v_C' = v_C - v_B = -15 m/s - (+5 m/s) = -20 m/s[/tex]

and the direction will be westward (negative sign).

Answer:

56 m/s

Explanation:

The time we are considering is

t = 15 s

The vertical velocity of the projectile is given by

[tex]v_y(t) = v_{0y}-gt[/tex]

where

[tex]v_{0y}=100 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

Substituting t=15 s, we find the vertical velocity of the projectile at that time:

[tex]v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s[/tex]

where the negative sign means the direction is now downward.

The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:

[tex]v_x = 30 m/s[/tex]

So, the magnitude of the velocity at the moment of impact is

[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s[/tex]

Answer:

[tex]2.64\cdot 10^5 m/s[/tex], the speed does not change

Explanation:

The magnetic force on the electron is equal to the product between its mass and its acceleration:

[tex]qvB = ma[/tex]

where

q is the electron charge

v is the electron speed

B = 0.14 T is the magnetic field

m is the electron's mass

[tex]a=6.5\cdot 10^{15}m/s^2[/tex] is the acceleration (centripetal acceleration)

Solving for v, we find

[tex]v=\frac{ma}{qB}=\frac{(9.11\cdot 10^{-31} kg)(6.5\cdot 10^{15} m/s^2)}{(1.6\cdot 10^{-19} C)(0.14 T)}=2.64\cdot 10^5 m/s[/tex]

The speed of the electron does not change, because the acceleration is a centripetal acceleration, so it acts perpendicular to the direction of motion of the electron; therefore, no work is done on the electron by the magnetic force, and therefore the electron does not gain kinetic energy, which means that its speed does not change.

Answer:

[tex]2.54\cdot 10^{-34}m[/tex]

Explanation:

First of all, we need to find the final velocity of the ball just before it reaches the ground. Since the ball is in free-fall motion, its final velocity is given by

[tex]v^2 = u^2 +2gh[/tex]

where

v is the final velocity

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration due to gravity

h = 39.8 m is the height of the building

Solving for v,

[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(39.8 m)}=27.9 m/s[/tex]

Now we can calculate the ball's momentum:

[tex]p=mv=(0.0935 kg)(27.9 m/s)=2.61 kg m/s[/tex]

And now we can calculate the De Broglie's wavelength of the ball:

[tex]\lambda = \frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{2.61 kg m/s}=2.54\cdot 10^{-34}m[/tex]

Answer:

8.86 m

Explanation:

According to the law of conservation of energy, the elastic potential energy initially stored in the spring will be converted into gravitational potential energy of the block when it is at its maximum height:

[tex]\frac{1}{2}kx^2 = mgh[/tex]

where

k = 5100 N/m is the spring constant

x = 0.093 m is the spring compression

m = 0.254 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

h is the maximum height of the block

Solving the equation for h, we find

[tex]h=\frac{kx^2}{2mg}=\frac{(5100 N/m)(0.093 m)^2}{2(0.254 kg)(9.8 m/s^2)}=8.86 m[/tex]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

[tex]\Phi = EA cos \theta[/tex]

where

E is the electric field

A is the cross-sectional area

[tex]\theta[/tex] is the angle between the direction of the electric field and the normal to A

The flux is maximum when [tex]\theta=0^{\circ}[/tex], so we are in this situation and therefore [tex]cos \theta =1[/tex], so we can write

[tex]\Phi = EA[/tex]

Here we have:

[tex]\Phi = 7.50\cdot 10^5 N/m^2 C[/tex] is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

[tex]A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2[/tex]

And so, we can find the magnitude of the electric field:

[tex]E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C[/tex]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

[tex]a_c =w^2r = \frac{v^2}{r}[/tex]

in this case we know the speed of the runner

[tex]v =8.00\ m/s[/tex]

The radius "r" will be the distance from the runner to the center of the track

[tex]r = 28.2\ m[/tex]

[tex]a_c = \frac{8^2}{28.2}\ m/s^2[/tex]

[tex]a_c = 2.27\ m/s^2[/tex]

The answer is the last option

Answer:

5120 m/s

Explanation:

The acceleration due to gravity is:

g = MG / r²

where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.

Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.

g = MG / (h + R)²

Acceleration is the derivative of velocity:

dv/dt = MG / (h + R)²

Using chain rule, we can say:

(dv/dh) (dh/dt) = MG / (h + R)²

(dv/dh) v = MG / (h + R)²

Separate the variables:

v dv = MG / (h + R)² dh

Integrating:

∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²

½ v² |₀ᵛ = -MG / (h + R) |₀ʰ

½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)

½ v² = -MG / (h + R) + MG / R

½ v² = MGh / (R(h + R))

v² = 2MGh / (R(h + R))

Given:

M = 5.98×10²⁴ kg

R = 6.37×10⁶ m

h = 1.69×10⁶ m

G = 6.67×10⁻¹¹ m³/kg/s²

Plugging in:

v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))

v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷

v ≈ 5120 m/s

Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)

v ≈ 5760 m/s

So we know that our calculated velocity of 5120 m/s is a reasonable answer.

The Newton's second law  and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:

The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.

           [tex]F= - G \frac{Mm}{r^2}[/tex]

Where M and m are the mass of the two bodies and r is the distance.

Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.

Newton's second law establishes a relationship between force, mass and the acceleration of bodies.

           F = m a  

          [tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]

The distance of the body from the center of the planet is

          r = R + h

The acceleration is defined as the variation of velocity with time.

          [tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]

Let's use the chain rule

         [tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]  

The velocity is the derivative of the position with respect to the time.

         [tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]    

To solve we use the method of separation of variables and we integrate.

       [tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]

      [tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]  

Let's calculate

      v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]

      v = 5120 m/s

In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;

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Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

[tex]U=\frac{1}{2}k (\Delta x)^2[/tex]

where

k is the spring constant

[tex]\Delta x[/tex] is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

[tex]\Delta x = d-0=d[/tex]

and the amount of energy required is

[tex]U=\frac{1}{2}k d^2[/tex]

In the second case, the spring is compressed from x=0 to x=-d, so

[tex]\Delta x = -d -0 = -d[/tex]

and the amount of energy required is

[tex]U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2[/tex]

so we see that the amount of energy required is the same.

Answer:

A transverse wave is one in which the disturbance is parallel to the direction of travel

Explanation:

There are two types of waves:

- Transverse waves: in transverse waves, the disturbance (oscillation) occurs in a plane perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the disturbance (oscillation) occurs parallel to the direction of propagation of the wave

Therefore, we immediately see that the statement:

"A transverse wave is one in which the disturbance is parallel to the direction of travel"

is wrong, because it is actually the opposite: in a transverse wave, the disturbance is perpendicular to the direction of travel.

The false statement is that a transverse wave has a disturbance parallel to its direction of travel. A transverse wave has the disturbance perpendicular to its direction of travel, not parallel. An example of this would be the waves on a stringed instrument, contrasted with sound waves which are longitudinal (disturbance and wave travel in the same direction).

The false statement among the given options is: "A transverse wave is one in which the disturbance is parallel to the direction of travel." This is incorrect because in a transverse wave, the disturbance or oscillation is perpendicular to the direction of wave propagation. For example, the movement of a stringed instrument creates transverse waves, where the strings vibrate upwards and downwards while the wave travels horizontally along the string.

On the other hand, in a longitudinal wave, the disturbance is parallel to the direction of wave propagation. An example would be sound waves, where the air particles vibrate back and forth, in the same direction that the wave propagates.

It's also correct that waves can have both transverse and longitudinal components, such as seismic waves from earthquakes. Furthermore, a wave does carry energy from one location to another but does not result in the bulk movement of the material medium.

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(a) [tex]9.66\cdot 10^{-15} W/m^2[/tex]

First of all, we need to find the altitude of the satellite. The gravitational attraction between the Earth and the satellite is equal to the centripetal force that keeps the satellite in circular motion, so

[tex]G\frac{mM}{r^2}=m\omega^2 r[/tex] (1)

where

G is the gravitational constant

m is the satellite's mass

M is the earth's mass

r is the distance of the satellite from the Earth's centre

[tex]\omega[/tex] is the angular frequency of the satellite

The satellite here makes two orbits around the Earth per day, so its frequency is

[tex]\omega = \frac {2 \frac{rev}{day}}{24 \frac{h}{day} \cdot 60 \frac{min}{h}} \cdot \frac{2\pi rad/rev}{s/min}=1.45\cdot 10^{-4} rad/s[/tex]

And by solving eq.(1) for r, we find

[tex]r=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)}{(1.45\cdot 10^{-4} rad/s)^2}}=2.67\cdot 10^{7} m[/tex]

The radius of the Earth is

[tex]R=6.37\cdot 10^6 m[/tex]

So the altitude of the satellite is

[tex]h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6m=2.03\cdot 10^7 m[/tex]

The average intensity received by a GPS receiver on the Earth will be given by

[tex]I=\frac{P}{A}[/tex]

where

P = 50.0 W is the power

A is the area of a hemisphere, which is:

[tex]A=4\pi h^2 = 4 \pi (2.03\cdot 10^7 m)^2=5.18\cdot 10^{15} m^2[/tex]

So the intensity is

[tex]I=\frac{50.0 W}{5.18\cdot 10^{15}m^2}=9.66\cdot 10^{-15} W/m^2[/tex]

(b) [tex]2.70\cdot 10^{-6} V/m, 9.0\cdot 10^{-15}T[/tex], 0.068 s

The relationship between average intensity of an electromagnetic wave and amplitude of the electric field is

[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

E is the amplitude of the electric field

Solving for E, we find

[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(9.66\cdot 10^{-15} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=2.70\cdot 10^{-6} V/m[/tex]

Instead, the amplitude of the magnetic field is given by:

[tex]B=\frac{E}{c}=\frac{2.70\cdot 10^{-6} V/m}{3\cdot 10^8 m/s}=9.0\cdot 10^{-15}T[/tex]

The signal travels at the speed of light, so the time it takes to reach the Earth is the distance covered divided by the speed of light:

[tex]t=\frac{h}{c}=\frac{2.03\cdot 10^7 m}{3\cdot 10^8 m/s}=0.068 s[/tex]

(c) [tex]3.22\cdot 10^{-23}Pa[/tex]

In case of a perfect absorber (as in this case), the radiation pressure exerted by an electromagnetic wave on a surface is given by

[tex]p=\frac{I}{c}[/tex]

where

I is the average intensity

c is the speed of light

In this case, we have

[tex]I=9.66\cdot 10^{-15} W/m^2[/tex]

So the average pressure is

[tex]p=\frac{9.66\cdot 10^{-15} W/m^2}{3\cdot 10^8 m/s}=3.22\cdot 10^{-23}Pa[/tex]

(d) 0.190 m

The wavelength of the receiver must be tuned to the same wavelength as the transmitter (the satellite), which is given by

[tex]\lambda=\frac{c}{f}[/tex]

where

c is the speed of light

f is the frequency of the signal

For the satellite in the problem, the frequency is

[tex]f=1575.42 MHz=1575.42\cdot 10^6 Hz[/tex]

So the wavelength of the signal is:

[tex]\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42 \cdot 10^6 Hz}=0.190 m[/tex]

Answer:

2780 N

Explanation:

Pressure is defined as the ratio between the force applied and the area of the surface:

[tex]p=\frac{F}{A}[/tex]

Here we know the pressure:

[tex]p=2.63 \cdot 10^9 N/m^2[/tex]

we also know the diameter of the tip, d = 1.15 mm, so we can calculate the radius

[tex]r=\frac{1.15 mm}{2}=0.58 mm = 5.8\cdot 10^{-4} m[/tex]

and so the area

[tex]A=\pi r^2 = \pi (5.8\cdot 10^{-4} m)^2=1.057\cdot 10^{-6} m^2[/tex]

And so we can re-arrange the equation to find the force:

[tex]F=pA=(2.63\cdot 10^9 N/m^2)(1.057\cdot 10^{-6} m^2)=2780 N[/tex]

We can calculate the necessary force to drive the nail home by first calculating the area of the nail tip using its given diameter then using the given pressure and the formula for pressure (P = F/A) to solve for the force.

To calculate the force required to drive the nail we need to use the equation for pressure which is pressure = force/area. Pressure (P) is given as 2.63 x 109 N/m2. The area (A) can be calculated using the formula for the area of a circle which is A = πr2. The radius (r) of the nail tip can be calculated from its diameter (1.15 mm divided by 2) making sure to convert the units to meters.

After you derive the area, you will use it to calculate the force (F). Rearranging the formula for pressure to solve for force gives F = P * A. This will give the force necessary to drive the nail home exerting the stated pressure.

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Final Answer:

The magnitude of the block's velocity just after impact is 0.008 times the initial velocity of the bullet. The initial speed of the bullet is 1.25 m/s.

Explanation:

To solve this problem, we can use the principle of conservation of momentum. The mass of the bullet is 8.00 g, or 0.008 kg, and the mass of the block is 0.992 kg. Let's denote the velocity of the block just after impact as V and the initial velocity of the bullet as u. The equation for conservation of momentum is:

m_bullet * u = (m_bullet + m_block) * V

Substituting the given values, we have:

0.008 kg * u = (0.008 kg + 0.992 kg) * V

0.008 u = 1 * V

Now, let's find the magnitude of the block's velocity just after impact:

V = 0.008 u

Given that the impact compresses the spring 15.0 cm, which requires a force of 0.750 N, we can use Hooke's Law to find the spring constant:

Force = spring constant * displacement

0.750 N = k * 0.150 m

k = 5 N/m

Now, to find the initial speed of the bullet, we can use the principle of conservation of mechanical energy. The energy stored in the spring when compressed 0.250 cm is given as 5.0 J. Let's denote the initial speed of the bullet as v_i. The equation for conservation of mechanical energy is:

0.5 * k * (0.250 m)^2 = 0.5 * m_bullet * (v_i)^2

Substituting the given values, we have:

0.5 * 5 N/m * (0.0025 m^2) = 0.5 * 0.008 kg * (v_i)^2

0.00625 J = 0.004 kg * (v_i)^2

(v_i)^2 = 0.00625 J / 0.004 kg

(v_i)^2 = 1.5625 m^2/s^2

v_i = 1.25 m/s

a) The final velocity of the block is approximately 0.0000808 * v_bullet.

b) The initial speed of the bullet is approximately 3.68 m/s.

a) Final Velocity of the Block (v_final):

Given data:

Mass of the bullet (m_bullet): 8.00 g (convert to kg: 0.008 kg)

Mass of the block (m_block): 0.992 kg

Compression of the spring (x): 15.0 cm (convert to meters: 0.15 m)

Initial velocity of the bullet (v_bullet): to be determined

Using the conservation of linear momentum:

m_bullet * v_bullet = (m_bullet + m_block) * v_final

0.008 kg * v_bullet = (0.008 kg + 0.992 kg) * v_final

Solving for v_final:

v_final = (0.008 kg * v_bullet) / (1 kg + 0.992 kg)

v_final ≈ 0.0000808 * v_bullet

b) Initial Speed of the Bullet (v_bullet):

Given data:

Force required to compress the spring (F): 0.750 N

Compression of the spring (x): 0.250 cm (convert to meters: 0.0025 m)

Spring constant (k): to be determined

The work done in compressing the spring is given by W = (1/2) * k * x^2, and this work is equal to the initial kinetic energy (KE_initial) of the bullet:

KE_initial = W = (1/2) * k * x^2

The kinetic energy is also related to the initial speed of the bullet:

KE_initial = (1/2) * m_bullet * v_bullet^2

Setting the two expressions for kinetic energy equal to each other:

(1/2) * k * x^2 = (1/2) * m_bullet * v_bullet^2

Solving for v_bullet:

v_bullet = sqrt((k * x^2) / m_bullet)

Now, using the given information that F = kx, where F is the force required to compress the spring:

k = F / x

Substitute this value of k back into the equation for v_bullet:

v_bullet = sqrt((F * x) / m_bullet)

Substitute the known values:

v_bullet = sqrt((0.750 N * 0.0025 m) / 0.008 kg)

v_bullet ≈ 3.68 m/s

Answer:

Explanation:

The first one is false.  A vector has a magnitude and a direction; a scalar only has a magnitude.  The two cannot be added together.

The second one is false.  The magnitude of a vector is found using Pythagorean theorem: c² = a² + b².  The only way the magnitude of a vector (c) can be 0 is if both components are 0 (a=0 and b=0).

The third one is true.  A vector in Quadrant III will have negative components but can still have a positive magnitude.  For example, a vector with magnitude 1 and direction 225° has a positive magnitude and negative components.

The fourth one is false.  Rotating a vector will change it.

The fifth one is false.  A vector sum can only be 0 if the two vectors have equal magnitudes and opposite directions.

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

Answer:

3. Madison, WI

4. Quito, Equador

6. Main Sequence stars are becoming red giants

Explanation:

Answer:

1) The angle at which spider-man points his web shooter

2) The distance spider-man’s web shoots

3) Madison, WI

4) Quito, Equador

5) Upper left

6) Main sequence stars are becoming red giants.

Explanation:

1) The experimenters are choosing the angle at which to point the web shooter. This means the angle at which they shoot is independent of the other variables.

2) The distance the web shoots is not chosen by the experimenters and depends on the angle at which it is shot.

3) July is the 7th month and the graph for Madison, WI shows temperatures exceeding all other locations at this time.

4) Fluctuations refer to a changing of temperatures over seasons. Since Quito, Equador has a flat graph, this indicates no fluctuations.

5) This graph plots the hottest (blue) stars on the left and the highest luminosity (bright) stars at the top. Thus, bright blue stars are plotted in the top left.

6) At 5 billion years there are 0 giants and 100 main sequence stars. At 10 billions years there are 94 giants and 6 main sequence stars. This shows that the main sequence stars are becoming giants.

A) 1296.3 m

The initial velocity of the projectile is

[tex]u = 165 m/s[/tex]

and the angle is

[tex]\theta=75^{\circ}[/tex]

So, we can find the initial vertical velocity of the projectile, which is given by

[tex]u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s[/tex]

The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration

[tex]g=-9.8 m/s^2[/tex]

where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation

[tex]v_y ^2 - u_y^2 = 2gh[/tex]

where

[tex]v_y = 0[/tex] is the final vertical velocity

h is the maximum height

Solving for h, we find

[tex]h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m[/tex]

B) 32.5 s

In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.

We can find it by analyzing the vertical motion only. The vertical position at time t is given by

[tex]y(t) = u_y t + \frac{1}{2}gt^2[/tex]

By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:

[tex]0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)[/tex]

which has two solutions:

t = 0 --> beginning of the motion

[tex]u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s[/tex]

C) 1387.8 m

The range of the projectile can be found by analyzing the horizontal motion only.

In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:

[tex]u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s[/tex]

So, the horizontal position at time t is given by

[tex]x(t) = u_x t[/tex]

If we substitute

t = 32.5 s

which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.

So we have:

[tex]x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m[/tex]

D) 165 m/s

The speed of the projectile consists of two independent components:

- The horizontal velocity, which is constant during the motion, and which is equal to

[tex]v_x = u_x = 42.7 m/s[/tex]

- The vertical velocity, which changes during the motion, given by

[tex]v_y = u_y + gt[/tex]

Substituting

[tex]u_y = 159.4 m/s[/tex]

and

[tex]t=32.5 s[/tex]

We find the vertical velocity when the projectile reaches the ground

[tex]v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s[/tex]

which is the same as the initial vertical velocity, but with opposite direction.

Now that we have the two components, we can calculate the actual speed of the projectile:

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s[/tex]

and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.

E) [tex]-75^{\circ}[/tex]

The angle of impact is given by

[tex]\theta = tan^{-1} (\frac{|v_y|}{v_x})[/tex]

where

[tex]|v_y| = 159.4 m/s[/tex] is the final vertical velocity

[tex]v_x = 42.7 m/s[/tex] is the final horizontal velocity

We have taken the absolute value of [tex]v_y[/tex] since [tex]v_y[/tex] is negative; this means that the resulting angle will be BELOW the horizontal. So we have:

[tex]\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}[/tex]

which means [tex]-75.0^{\circ}[/tex], below the horizontal.

Final answer:

To determine the frequency ranges for UVA and UVB, we use the equation f = c/λ. The UVA frequency range is 7.5 × 10^14 Hz to 9.375 × 10^14 Hz, and the UVB frequency range is 9.375 × 10^14 Hz to 1.0345 × 10^15 Hz.

Explanation:

The frequency of electromagnetic radiation, including UV radiation, can be calculated using the formula c = λf, where c is the speed of light (approximately 3 × 10^8 m/s), λ is the wavelength, and f is the frequency. To find the frequency ranges for UVA and UVB, we can rearrange the equation to f = c/λ.

Calculating UVA Frequency

For UVA with a wavelength range of 320-400 nm (or 3.2 × 10^-7 m - 4 × 10^-7 m), we use the formula to calculate the frequency as:

• Lower frequency limit: f = (3 × 10^8 m/s) / (4 × 10^-7 m) = 7.5 × 10^14 Hz

• Upper frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz

Calculating UVB Frequency

For UVB with a wavelength range of 290-320 nm (or 2.9 × 10^-7 m - 3.2 × 10^-7 m), the frequency range is:

• Lower frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz

• Upper frequency limit: f = (3 × 10^8 m/s) / (2.9 × 10^-7 m) = 1.0345 × 10^15 Hz

The answer is:

The time took for the rock to reach its maximum height is 0.110 seconds.

In order to calculate the time needed for the rock to reach its maximum height, we need to calculate the initial vertical speed.

From the statement we know that the rock was launched at an initial speed of 2.1 m/s at an angle of 30° above the horizontal, so, calculating we have:

[tex]V_x=2.1\frac{m}{s} *cos(30\°)=1.82\frac{m}{s} \\\\V_y=2.1\frac{m}{s} *sin(30\°)=1.05\frac{m}{s}[/tex]

Also, we know that at the maximum height, the speeds tends to 0. So, using the following equation and substituting "v" equal to 0, we have:

[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]

Where,

t is the time in seconds.

V is the initial speed

g is the gravity acceleration.

Using gravity acceleration equal to [tex]9.81\frac{m}{s^{2} }[/tex]  we have:

[tex]V=V_o-g*t\\\\0=V_o-g*t\\\\g*t=V_o\\\\t=\frac{V_o}{g}[/tex]

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

[tex]t=\frac{1.05\frac{m}{s}}{9.81\frac{m}{s^{2}}}=0.1070seconds=0.110seconds[/tex]

Hence, the correct option is the last option, the time took for the rock to reach its maximum height is 0.110 seconds.

Have a nice day!

Answer:

All of the choices are correct

Explanation:

The power dissipated in a single resistor connected to a battery is given by:

[tex]P=VI = I^2 R=\frac{V^2}{R}[/tex]

where

V is the voltage

I is the current

R is the resistance

Let's analyze each case:

A) Doubling the voltage (V'=2V) and reducing the current by a factor of 2 (I'=I/2). The new power dissipated is:

[tex]P'=V'I'=(2V)(\frac{I}{2})=VI=P[/tex] --> the power is unchanged

B) Doubling the voltage (V'=2V) and increasing the resistance by a factor of 4 (R'=4R). The new power dissipated is:

[tex]P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}[/tex] --> the power is unchanged

C) Doubling the current (I'=2I) and reducing the resistance by a factor of four (R'=R/4). The new power dissipated is:

[tex]P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R[/tex] --> the power is unchanged

Final answer:

Option (B), which involves doubling the voltage and increasing the resistance by a factor of four, is the correct scenario where the electric power dissipated in the resistor remains unchanged, explained by the formulas P = V^2/R and P = I^2R.

Explanation:

The question addresses how voltage, current, and resistance affect electric power dissipation in a resistor. Electric power (P) dissipated by a resistor can be calculated using the formula P = V2/R or P = I2R, where V is the voltage across the resistor, I is the current through the resistor, and R is the resistance. To keep the power dissipated unchanged, we need to ensure that any changes in voltage, current, or resistance occur in a way that does not change the final calculation of P.

Option (A) suggests doubling the voltage and reducing the current by a factor of two. This will not keep power dissipated the same due to the direct relationship in the formula P = I2R.

Option (B), doubling the voltage and increasing the resistance by a factor of four, will keep the power dissipated in the resistor unchanged because the power can also be expressed as P = V2/R, and thus doubling V while quadrupling R does not change the P.

Option (C) involves doubling the current and reducing the resistance by a factor of four. According to P = I2R, this change will not keep the power dissipated unchanged as the increase in current will significantly raise power dissipation due to its squared relationship in the formula.

Therefore, Option (B) correctly describes a method to change voltage and resistance in such a way that the power dissipated in the resistor remains unchanged.

Explanation:

We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:

[tex]c=\lambda.\nu[/tex]   (1)

Clearing [tex]\lambda[/tex] :

[tex]\lambda=\frac{c}{\nu}[/tex]   (2)

Knowing [tex]\nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}[/tex]

[tex]\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}[/tex]   (3)

[tex]\lambda=136.363m[/tex] This is the wavelength of the radiation produced by the microwave

The wavelength of the radiation produced by a microwave oven operating at 2.20 GHz is approximately 13.6 cm.

Calculating the Wavelength of Microwave Oven Radiation

To find the wavelength of the radiation produced by a microwave oven that operates at 2.20 GHz, we use the formula c = λν, where c is the speed of light (3.00 × 10¸ m/s), λ is the wavelength in meters, and ν is the frequency in hertz (Hz).

First, we convert the frequency from gigahertz (GHz) to hertz (Hz) by multiplying it by 10¹:
2.20 GHz × 10¹ = 2.20 × 10¹ Hz.

Next, we rearrange the formula to solve for λ:

λ = c / ν

Now, we plug in the values:

λ = (3.00 × 10¸ m/s) / (2.20 × 10¹ Hz)

Calculating this gives:

λ = (3.00 × 10¸) / (2.20 × 10¹)

λ = 1.36 × 10² m

Therefore, the wavelength of the radiation emitted by the microwave oven is approximately 13.6 cm.