Answer:
696.83 m/s
Explanation:
m = mass of water = 565 g = 0.565 kg
c = specific heat of water = 4186 J/(kg⁰C)
ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C
v = speed gained by water
Using conservation of energy
Kinetic energy gained by water = heat required to warm water
(0.5) m v² = m c ΔT
(0.5) v² = c ΔT
(0.5) v² = (4186) (58)
v = 696.83 m/s
106 m/s in a uniform 1.9 x 105 N/C electric field. The field accelerates the Problem 6: An electron has an initial velocity of 5.25 electron in the direction opposite to its initial velocity. Part (a) What is the direction of the electric field? MultipleChoice 1) The field is in the direction of the electron's initial velocity 2) The field is in the direction to the right of the clectron's initial velocity 3) The ficld is in the opposite dircction of the elcctron's initial velocity 4) The field is in another direction not listed here Part (b) How far does the electron travel before coming to rest in m? Numeric : A numeric value is expected and not an expression Part (c) How long does it take the clectron to come to rest in s? Numeric A numeric value is expected and not an expression. Part (d) What is the magnitude of the electron's velocity (in m/s) when it returns to its starting point in the opposite direction of its initial velocity? Numeric : A numeric value is expected and not an expression
(a) 1) The field is in the direction of the electron's initial velocity
The electric field is in a direction opposite to the initial velocity of the electron.
Let's remind that, when an electric charge is immersed in an electric field:
- if the charge is positive, the charge experiences a force in the same direction as the electric field direction
- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction
In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.
(b) [tex]4.13\cdot 10^{-4} m[/tex]
We have:
Electron's initial velocity: [tex]u=5.25\cdot 10^6 m/s[/tex]
Electric field magnitude: [tex]E=1.9 \cdot 10^5 N/C[/tex]
Electron charge: [tex]q=-1.6\cdot 10^{-19} C[/tex]
Mass of the electron: [tex]m=9.11\cdot 10^{-31}kg[/tex]
The electric force exerted on the electron is:
[tex]F=qE=(-1.6\cdot 10^{-19} C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^{-14}N[/tex] (the negative sign means the direction of the force is opposite to its initial velocity)
The electron's acceleration is given by:
[tex]a=\frac{F}{m}=\frac{3.04\cdot 10^{-14} N}{9.11\cdot 10^{-31} kg}=-3.34\cdot 10^{16} m/s^2[/tex]
Now we can use the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final speed (the electron comes to rest)
d is the total distance travelled by the electron
Solving for d,
[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(5.25\cdot 10^6 m/s)^2}{2(-3.34\cdot 10^{16} m/s^2)}=4.13\cdot 10^{-4} m[/tex]
(c) [tex]1.57\cdot 10^{-10}s[/tex]
We can use the following equation:
[tex]a=\frac{v-u}{t}[/tex]
where we have
[tex]a=-3.34\cdot 10^{16}m/s^2[/tex] is the electron's acceleration
v = 0 is its final speed
[tex]u=5.25\cdot 10^6 m/s[/tex] is the initial speed
t is the time it takes for the electron to come at rest
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{0-(5.25\cdot 10^6 m/s)}{-3.34\cdot 10^{16} m/s^2}=1.57\cdot 10^{-10}s[/tex]
(d) [tex]5.25\cdot 10^6 m/s[/tex]
This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.
So we have:
u = 0 is the initial speed of the electron
[tex]a=3.34\cdot 10^{16}m/s^2[/tex]
[tex]d=4.13\cdot 10^{-4} m[/tex] is the distance covered to go back
So we can use the following equation:
[tex]v^2 - u^2 = 2ad[/tex]
to find v, the new final speed:
[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2 + 2(3.34\cdot 10^{16} m/s^2)(4.13\cdot 10^{-4} m)}=5.25\cdot 10^6 m/s[/tex]
Select the best definition for wavelength. the height of an oscillating electromagnetic wave the rate at which electromagnetic waves oscillate. the distance between two crests of an electromagnetic wave. the oscillations of electric and magnetic fields Select the best definition for frequency. the oscillations of electric and magnetic fields. the height of an oscillating electromagnetic wave. the distance between two crests of an electromagnetic wave. the rate at which electromagnetic waves oscillate
Answer:
Wavelength: the distance between two crests of an electromagnetic wave
Frequency: the rate at which electromagnetic waves oscillate
Explanation:
For a transverse wave (such as the electromagnetic waves), we can define the following quantities:
Wavelength: it corresponds to the distance between two consecutive crests (or between two consecutive troughs) of the wave
Frequency: the number of complete oscillations of the wave per unit time
Period: it is the time it takes for the wave to complete one oscillation - it is equal to the reciprocal of the frequency
Amplitude: it is the maximum displacement of the wave, measured with respect to its equilibrium position
A bearing is designed to ____ A reduce friction B. support a load C.guide moving parts such as wheels, shafts and pivots D. all of the above
Answer:
Option (A)
Explanation:
A ball bearing is a device which is use to reduce the friction.
The outer rim of the bearing is fixed with the part of machine and inner rim is fitted into shafts. Now the shafts rotates and only the small spheres in the bearing will rotate. The friction can be further reduced by apply the oil or grease to the bearing.
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.67 kg and 2.59 kg, and the length of the wire is 1.01 m. Find the velocity of the ball just after the collision.
Answer:
0.96 m/s
Explanation:
Consider the motion of the ball before collision with the block
h = height from which the ball is dropped = length of the wire = 1.01 m
m = mass of the ball = 1.67 kg
v = speed of the ball just before collision with the block = ?
Using conservation of energy for the ball
Kinetic energy of the ball at the bottom = Potential energy of the ball at the top
(0.5) m v² = mgh
(0.5) v² = gh
(0.5) v² = (9.8) (1.01)
v = 4.45 m/s
consider the collision between the ball and the block :
m = mass of the ball = 1.67 kg
v = velocity of the ball just before collision with the block = 4.45 m/s
v' = velocity of the ball just after collision with the block
M = mass of the block = 2.59 kg
V = velocity of the block just before collision with the ball = 0 m/s
V' = velocity of the block just after collision with the ball
Using conservation of momentum
mv + MV = mv' + MV'
(1.67) (4.45) + (2.59) (0) = 1.67 v' + (2.59) V'
7.43 = 1.67 v' + (2.59) V'
[tex]V' = \frac{(7.43 - 1.67 v')}{2.59}[/tex] eq-1
Using conservation of kinetic energy
(0.5) mv² + (0.5) MV² = (0.5) mv'² + (0.5) MV'²
mv² + MV² = mv'² + MV'²
(1.67) (4.45)² + (2.59) (0)² = 1.67 v'² + (2.59) V'²
using eq-1
(1.67) (4.45)² = 1.67 v'² + (2.59) ((7.43 - 1.67 v')/2.59)²
v' = - 0.96 m/s
the negative sign indicates the direction which is opposite to its direction before colliding with the block.
To find the velocity of the ball just after the elastic collision, apply the formula for one-dimensional elastic collision. Given that the ball initially fell from a certain height, calculate its initial velocity using the conservation of energy principle. Insert these results, along with the given masses, into the elastic collision formula to achieve the answer.
Explanation:This question explores the principles of conservation of momentum and the properties of an elastic collision. Given that this collision is perfectly elastic, we can use the formula for an elastic collision in one-dimension:
v1' = ((m1 - m2) / (m1 + m2)) * v1 + ((2 * m2) / (m1 + m2)) * v2
where v1' is the velocity of the ball after collision, m1 and m2 are the masses of the ball and the block respectively, v1 and v2 are the velocities of the ball and block before the collision. The ball was initially dropped from a height, so we'd first need to calculate its velocity just before collision using the conservation of energy principle:
v1 = sqrt(2* g * h)
where g is the acceleration due to gravity and h is the height (length of the wire). After figuring out the value of v1, we just need to plug the values into the first equation to find v1' - velocity of the ball after collision.
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A 920 kg cannon fires a 3.5 kg shell at initial acceleration of 95 m/s^2. What is the cannon's recoil force?
Answer:
The cannon recoils with a force of 332.5 N
Explanation:
By Newton's third law Recoil force on cannon = Force in shell.
Force in shell = Mass of shell x Acceleration of shell
Mass of shell = 3.5 kg
Acceleration of shell = 95 m/s²
Force in shell = 3.5 x 95 = 332.5 N
Recoil force on cannon = 332.5 N
So, the cannon recoils with a force of 332.5 N
One of the fastest recorded pitches in major-league baseball, thrown by Tim Lincecum in 2009, was clocked at 101.0 mi/h. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 ft away?
Final answer:
A baseball pitched horizontally at 101.0 mi/h will travel 60.5 feet to home plate in about 0.408 seconds; during that time, it will fall approximately 2.68 feet due to gravity.
Explanation:
To calculate how far a baseball would fall vertically when thrown horizontally at 101.0 mi/h, we can divide the problem into two separate motions - horizontal and vertical. Since the vertical and horizontal motions are independent, we can analyze them separately. First, we need to convert the speed to feet per second (1 mi/h = 1.467 ft/s), multiply 101.0 mi/h by 1.467 to get the horizontal velocity in feet per second, and then use that to find the time it takes for the ball to travel the horizontal distance to home plate.
Horizontal distance to home plate = 60.5 ft. Horizontal speed Vh = 101.0 mi/h * 1.467 ft/s/mi/h = 148.137 ft/s. Time to reach home plate t = distance/speed = 60.5 ft / 148.137 ft/s = 0.408 s (approximately).
Next, to find out how far the ball falls vertically, we'll use the formula for the distance an object falls due to gravity: d = 0.5 * g * t2, where g is the acceleration due to gravity. In standard English units, g is approximately 32.2 ft/s2. Plugging in the numbers, we get d = 0.5 * 32.2 ft/s2 * (0.408 s)2 = 2.68 ft (approximately).
Thus, a baseball thrown horizontally at 101.0 mi/h will fall about 2.68 feet by the time it reaches home plate.
The correct answer is that the ball would fall vertically by approximately 2.63 ft by the time it reached home plate, 60.5 ft away.
To solve this problem, we can use the kinematic equation for the vertical motion of the ball under the influence of gravity, assuming no air resistance. The equation is:
[tex]\[ y = v_{0y} t + \frac{1}{2} g t^2 \][/tex]
where:
- y is the vertical displacement,
- [tex]\( v_{0y} \)[/tex] is the initial vertical velocity (which is 0 m/s in this case since the ball is thrown horizontally),
- g is the acceleration due to gravity (9.81 m/s, but we will use 32.2 ft/s for consistency with the imperial units given in the problem),
- t is the time of flight.
Since the initial vertical velocity [tex]\( v_{0y} \)[/tex] is 0, the equation simplifies to:
[tex]\[ y = \frac{1}{2} g t^2 \][/tex]
We need to find the time t it takes for the ball to travel 60.5 ft horizontally. The horizontal velocity [tex]\( v_x \)[/tex] is given as 101 mi/h. We convert this to feet per second:
[tex]\[ v_x = 101 \text{ mi/h} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \][/tex]
[tex]\[ v_x = 101 \times 5280 \times \frac{1}{3600} \text{ ft/s} \][/tex]
[tex]\[ v_x = 147.67 \text{ ft/s} \][/tex]
Now we can calculate the time t it takes to travel 60.5 ft:
[tex]\[ t = \frac{\text{distance}}{\text{velocity}} = \frac{60.5 \text{ ft}}{147.67 \text{ ft/s}} \][/tex]
[tex]\[ t \ = 0.409 \text{ s} \][/tex]
Using this time, we can find the vertical displacement y :
[tex]\[ y = \frac{1}{2} \times 32.2 \text{ ft/s}^2 \times (0.409 \text{ s})^2 \][/tex]
[tex]\[ y = \frac{1}{2} \times 32.2 \times 0.167 \text{ ft} \][/tex]
[tex]\[ y = 16.7 \times 0.167 \text{ ft} \][/tex]
[tex]\[ y \ = 2.63 \text{ ft} \][/tex]
An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds. Which describes the acceleration?
Answer:
4 m/s2 in negative acceleration
Explanation:
If An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds, it is 4 m/s2 in negative acceleration.
Hope this helps!
Answer:
[tex]-4\frac{m}{s^2}[/tex]
Explanation:
The object changes its speed over some time, this means that there is an acceleration.
It has a uniformly accelerated movement.
The Formula for finding speed in a uniformly accelerated motion is
[tex]a=\frac{V_{f}-V_{o}}{t}[/tex]
[tex]V_{o}= 65\frac{m}{s}\\V_{f}= 25\frac{m}{s} \\t= 10s[/tex]
Replace
[tex]a=\frac{(25-65)\frac{m}{s} }{10 s}\\ a=\frac{-40\frac{m}{s} }{10s}\\a= -4\frac{m}{s^2}[/tex]
Acceleration gives us a negative value this means that it is slowing.
A sample of blood is placed in a centrifuge of radius 12.0 cm. The mass of a red blood cell is 3.0 ✕ 10^−16 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 ✕ 10^−11 N. At how many revolutions per second should the centrifuge be operated?
Answer:
167.85 rev / s
Explanation:
r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N
F = m r w^2
where, w is the angular velocity.
4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2
w = 1054.1 rad / s
w = 2 π f
f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s
The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.
Recall :
Frequency, f = ω/2πForce, F = mω²rMass, m = [tex] 3 \times 10^{-16}[/tex] Force, F = [tex] 4 \times 10^{-11} N[/tex] Radius, r = 12 cm = 12/100 = 0.12 mWe calculate the angular velocity, ω thus :
ω² = F/mr
ω² = [tex] \frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}[/tex]
ω = [tex] \sqrt{1.11 \times 10^{6}} = 1053.56 rad/s[/tex]
Frequency = 1053.56 ÷ (2π)
Frequency = 167.68 Hz
Therefore, the Number of revolutions per seconds would be about 167.8 Hz
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A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed is 37.2 m/s. What is the magnitude of the acceleration (in m/s?) of the particle?
The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude
[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]
We have
[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]
where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so
[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]
We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so
[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]
Then the magnitude of the particle's acceleration is
[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]
The magnitude of the acceleration of the particle is approximately 70.55 m/s^2, calculated by using the formulas for combined radial and tangential acceleration in circular motion.
Explanation:In this physics problem, the particle not only moves around in a circle but is also experiencing an increase in speed which is a case of combined radial and tangential acceleration. Radial acceleration, known as centripetal acceleration (ar), is the result of the change in direction of the velocity vector, while tangential acceleration (at) comes from changes in speed.
The total acceleration of an object in circular motion is given by:
a = sqrt((ar^2) + (at^2))
Centripetal acceleration can be calculated using the formula ar = v^2 / r, where: v = speed (37.2 m/s), r = radius of the circle (21 m). This gives us ar = (37.2^2) / 21, which approximately equals 66.62 m/s^2.
The tangential acceleration is given in the problem: at = 23.1 m/s^2.
We therefore calculate the total acceleration using the formula above which gives us:
a = sqrt((66.62^2) + (23.1^2)) which approximately equals 70.55 m/s^2.
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How many electrons leave a 9.0V battery every minute if it is connected to a resistance of 1.4?? O 80x 1020 O 6.7x 1020 O 5.1 x 1021 O 24x 1021
Answer:
2.4 x 10^21
Explanation:
V = 9 V, R = 1.4 ohm, t = 1 minute = 60 second
Use Ohm's law
V = I R
I = V / R
I = 9 / 1.4
I = 6.43 A
Now use Q = I t
Q = 6.43 x 60 = 385.7 C
Number of electrons passing in 1 minute , n
= total charge in one minute / charge of one electron
n = 385.7 / (1.6 x 10^-19) = 2.4 x 10^21
A particle with a charge of -2.7 ?C and a mass of 3.8 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 36 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.
Answer:
0.0018 V
Explanation:
According to the law of conservation of energy, the kinetic energy gained by the particle is equal to the electric potential energy lost:
[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]
where
[tex]m=3.8\cdot 10^{-6} kg[/tex] is the mass of the particle
[tex]v=36 m/s[/tex] is the final speed of the particle
q = -2.7 C is the charge
[tex]\Delta V[/tex] is the potential difference between the two points
Solving for [tex]\Delta V[/tex], we find
[tex]\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V[/tex]
The particle has been accelerated by this potential difference: since it is a negative charge, it means that the particle has moved from a point at lower potential towards a point of higher potential.
So, since the initial point is A and the final point is B, the result is
[tex]V_B - V_A = 0.0018 V[/tex]
A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the car moves 11.5 m/s, both in the same direction. What is the mass of the car?
Answer:
Mass of car = 1098 kg
Explanation:
Here law of conservation of momentum is applied.
Let mass of car be m.
Initial momentum = Final momentum.
Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s
Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m
We have
19792.5+11.5m = 32416.5
m = 1097.97 kg
Mass of car = 1098 kg
Answer:1074.26
Explanation:just got it right on my accelus
A 7.5-N net force is applied to a 47-kg object. What is the object's acceleration?
Answer:
Object's acceleration, a = 0.15 m/s²
Explanation:
It is given that,
Net force acting on the object, F = 7.5 N
Mass of an object, m = 47 kg
We need to find the acceleration of the object. It can be calculated using second law of motion as :
F = ma
a = acceleration of the object
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{7.5\ N}{47\ kg}[/tex]
[tex]a=0.15\ m/s^2[/tex]
So, the acceleration of the object is 0.15 m/s². Hence, this is the required solution.
Sexual harassment is a form of sex discrimination and is against the law.
True
False
Answer:
True.
Explanation:
This statament is true.
A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300 kV at the power station and carries the current to a city located 156 km from the power station. What is the power loss due to resistance in the line? Answer in units of MW
[tex]\boxed{P=96.09MW}[/tex]
Explanation:First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:
[tex]R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega[/tex]
So we can calculate the power loss as:
[tex]P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}[/tex]
Finally, the power loss due to resistance in the line is 96.09MW
A reaction taking place in a container with a piston-cylinder assembly at constant temperature produces a gas, and the volume increases from 127 mL to 654 mL against an external pressure of 860 torr. Calculate the work done in Joules (J)
Answer:
60.4 J
Explanation:
The work done by the gas is given by:
[tex]W=p(V_f-V_i)[/tex]
where
p is the gas pressure
[tex]V_f[/tex] is the final volume of the gas
[tex]V_i[/tex] is the initial volume
We must convert all the quantities into SI units:
[tex]p=860 torr \cdot \frac{1.013\cdot 10^5 pa}{760 torr}=1.146\cdot 10^5 Pa[/tex]
[tex]V_i = 127 mL = 0.127 L = 0.127 dm^3 = 0.127 \cdot 10^{-3}m^2[/tex]
[tex]V_f = 654 mL = 0.654 L = 0.654 dm^3 = 0.654 \cdot 10^{-3}m^2[/tex]
So the work done is
[tex]W=(1.146\cdot 10^5 Pa)(0.654\cdot 10^{-3} m^3-0.127\cdot 10^{-3} m^3)=60.4 J[/tex]
Answer:
[tex]W=60.4 J[/tex]
Work done is 60.4 Joules (J)
Explanation:
Work done 'W' is given by:
[tex]W=P\triangle V[/tex]
Where:
ΔV is the change in Volume.
P is the pressure.
Change in Volume=Final Volume-Initial Volume
Initial Volume= 127 mL
Final Volume= 654 mL
ΔV=654-127 (mL)
ΔV=527 mL
ΔV=0.527 L
Pressure Conversion:
1 atm=760 torr
[tex]P=\frac{860}{760} \\P=1.1315\ atm[/tex]
Now,
[tex]W=P\triangle V[/tex]
[tex]W=1.1315\ atm*0.527\ L[/tex]
[tex]W=0.5963 L.atm[/tex]
In joule (J): (Conversion 1 atm. L=101.325 J)
[tex]W=0.5963\ L.atm*\frac{101.325 J}{1\ L.atm}[/tex]
[tex]W=60.4 J[/tex]
Work done is 60.4 Joules (J)
A 6 N force and a 15 N force act on an object. The moment arm of the 6 N force is 0.4 m. If the 15 N 20. force provides 5 times the torque of the 6 N force, what is its moment arm?
Answer:
0.8 m
Explanation:
For "6 N" force :
F = magnitude of the force = 6 N
r = moment arm = 0.4 m
Torque due to "6 N" force is given as
τ = r F
τ = (0.4) (6)
τ = 2.4 Nm
For " 15 N" force :
F' = magnitude of the force = 15 N
r' = moment arm = ?
τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm
Torque due to "15 N" force is given as
τ' = r' F'
12 = r' (15)
r' = 0.8 m
So the moment arm for "15 N" force is 0.8 m
A projectile is shot from the edge of a cliff 140 m above ground with an initial speed of 120 m/s at an angle of 38 degrees above the horizontal. What is the time taken by the projectile to hit the ground 140 m below the cliff?(g = 9.8 m/s²)
Answer:
17 seconds
Explanation:
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 = 140 + (120 sin 38) t + ½ (-9.8) t²
4.9 t² - 73.9 t - 140 = 0
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8
t = -1.7, 16.8
Since t can't be negative, t = 16.8. Rounding to 2 sig-figs, the projectile lands after 17 seconds.
An automobile engine has an efficiency of 19.0%. If it produces 23.0 kJ of mechanical work per second, the heat rejected per second from the engine is closest to: (a) 77.0 kJ (b) 81.5 kJ (c)86.5 kJ (d) 92.0 kJ (e) 98.1 kJ
Answer:
(e) 98,1 KJ
Explanation:
The engine produces 19%; it means, it rejects 81% of energy. ⇒ 81/19=4.26 times.
The engine produces 23 kJ; it means it rejects 23 * 4.26 = 98.05263 kJ
The emf of a battery is equal to its terminal potential difference A) under all conditions B) only when the battery is being charged C) only when a large current is in the battery D) only when there is no current in the battery E) under no conditions
D. The chemical potential energy of real battery decreases, it develops internal resistance and therefore the potential difference across its terminals decreases if its current increases.
An elbow has a loss coefficient K of 0.4. Considering a pipe with friction factor of 0.025 and diameter of 20 mm, what is the equivalent length (Le) for the elbow in mm?
Answer:
320 mm
Explanation:
The equivalent length for a minor loss is:
Le = K D / f
where K is the loss coefficient, D is the diameter, and f is the friction factor.
Le = (0.4) (20 mm) / 0.025
Le = 320 mm
The equivalent length (Le) for the elbow with a loss coefficient (K) of 0.4, a pipe with a friction factor of 0.025, and a diameter of 20 mm, is 320 mm.
To calculate the equivalent length (Le) for the elbow in terms of pipe diameter, we use the relationship between the loss coefficient (K), the friction factor (f), and the diameter (D) of the pipe, given by the equation Le = (K * D) / f. In this case, the loss coefficient for the elbow is 0.4, the friction factor is 0.025, and the diameter of the pipe is 20 mm. Plugging these values into the equation, we get:
Le = (0.4 * 20 mm) / 0.025 = 320 mm.
Therefore, the equivalent length for the elbow is 320 mm.
a person when asked to speak up,increases her sound level from 30dB to 60dB.The amount of power per unit area increased by? a)3000 times b)30 times c)2 times d)1000 times
Answer:
d) 1000 times
Explanation:
As we know that difference of sound level is given as
[tex]L_2 - L_1 = 10 Log \frac{I_2}{I_1}[/tex]
so here we need to find the ratio of two intensity
it is given as
[tex]Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}[/tex]
[tex]Log\frac{I_2}{I_1} = \frac{60 - 30}{10}[/tex]
[tex]Log\frac{I_2}{I_1} = 3[/tex]
now we have
[tex]\frac{I_2}{I_1} = 10^3[/tex]
so it is
d) 1000 times
Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizontal. a) How long will the ball be in the air? b) What maximum height will the ball attain? c) How far away from Tom will the ball land? d) What speed will the ball have in the instant just before it lands?
Answer:
(a) 2.34 s
(b) 6.71 m
(c) 38.35 m
(d) 20 m/s
Explanation:
u = 20 m/s, theta = 35 degree
(a) The formula for the time of flight is given by
[tex]T = \frac{2 u Sin\theta }{g}[/tex]
[tex]T = \frac{2 \times 20 \times Sin35 }{9.8}[/tex]
T = 2.34 second
(b) The formula for the maximum height is given by
[tex]H = \frac{u^{2} \times Sin^{2}\theta }{2g}[/tex]
[tex]H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}[/tex]
H = 6.71 m
(c) The formula for the range is given by
[tex]R = \frac{u^{2} \times Sin 2\theta }{g}[/tex]
[tex]R = \frac{20^{2} \times Sin 2 \times 35}{9.8}[/tex]
R = 38.35 m
(d) It hits with the same speed at the initial speed.
Three resistors are connected in series across a 15-V power supply. If the potential drops across resistors 1 and 2 are 4.1 volts and 3.1 volts, what is the exact potential drop (in volts) across resistor 3?
Answer:
7.8 V
Explanation:
When resistors are connected in series, they are connected in the same branch of the circuit. This means that the current flowing each resistor is the same, while the sum of the voltage drops across each resistor is equal to the potential difference of the power supply:
[tex]V= V_1 + V_2 + V_3[/tex]
In this circuit we have:
V = 15 V
[tex]V_1 = 4.1 V[/tex]
[tex]V_2 = 3.1 V[/tex]
So, the voltage drop across resistor 3 is
[tex]V_3 = V-V_1 - V_2 = 15 V - 4.1 V - 3.1 V=7.8 V[/tex]
Final answer:
The exact potential drop across the third resistor, when three resistors are connected in series to a 15-V power supply and the voltage drops across the first two are 4.1V and 3.1V, is 7.8 volts.
Explanation:
The question asks for the exact potential drop across resistor 3 when three resistors are connected in series across a 15-V power supply, and the potential drops across the first two resistors are known. In a series circuit, the total voltage supplied is equal to the sum of individual voltage drops across all components. Given that the voltage drops across resistors 1 and 2 are 4.1 volts and 3.1 volts respectively, the voltage drop across resistor 3 can be calculated by subtracting the sum of the voltage drops across resistors 1 and 2 from the total voltage supplied by the power source.
Therefore, the calculation is:
Total voltage supplied by the power source = 15 volts
Sum of voltage drops across resistor 1 and 2 = 4.1 volts + 3.1 volts = 7.2 volts
Voltage drop across resistor 3 = Total voltage - Sum of voltage drops across resistor 1 and 2 = 15 volts - 7.2 volts = 7.8 volts.
Two identical circular plates have a diameter of 8.2 cm. They are placed parallel to each other separated by a distance of 2.1 mm. What is the capacitance if paper, with a k-3.7, is placed between the plates? O 92 pF O 82 p O 62 pF ? 13oF
Answer:
82 pF
Explanation:
diameter = 8.2 cm, distance, d = 2.1 mm = 0.0021 m,
dielectric constant, k = 3.7
radius = 1/2 x diameter = 8.2 / 2 = 4.1 cm = 0.041 m
The formula for the capacitance of parallel plate capacitor is given by
[tex]C = \frac{k \varepsilon _{0}\times A}{d}[/tex]
[tex]C = \frac{3.7 \times 8.854\times 10^{-12} \times \pi \times 0.041 \times 0.041}{0.0021}[/tex]
C = 8.23 x 10 ^-11 F
C = 82 pF
A tank holds a 1.44-m thick layer of oil that floats on a 0.98-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.40 and 1.52, respectively. A ray originating at O crosses the brine-oil interface at a point 0.7 m from the axis. The ray continues and emerges into the air above the oil. What is the angle that the ray in the air makes with the vertical axis
Answer:
Angle of ray makes with the vertical is 62.1 degree
Explanation:
As per the ray diagram we know that the angle of incidence on oil brine interface will be given as
[tex]tan\theta_i = \frac{0.7}{0.98}[/tex]
[tex]\theta_i = 35.5^0[/tex]
now by Snell'a law at that interface we have
[tex]\mu_1 sin\theta_i = \mu_2 sin \theta_r[/tex]
now we will have
[tex]1.52 sin35.5 = 1.40 sin\theta_r[/tex]
[tex]\theta_r = 39.12^0[/tex]
now this is the angle of incidence for oil air interface
so now again by Snell's law we will have
[tex]\mu_2 sin\theta_i' = \mu_{air} sin\theta[/tex]
[tex]1.40 sin39.12 = 1 sin\theta[/tex]
[tex]\theta = 62.1^0[/tex]
The problem involves applying Snell's law twice, once for the brine-to-oil interface and then for the oil-to-air interface, with angles calculated from the normal lines. Severally, the use of the Pythagorean theorem gives the angle in brine, Snell's law then gives the angle in oil, and finally in the air. The desired angle is found as an augmentation to 90 degrees since it should be calculated from the axis, not the normal line.
Explanation:Firstly we need to calculate the angle that the ray makes with the norm or perpendicular line to the brine-oil interface. This can be done using the Pythagorean theorem, given that we have the height of the brine (0.98m), and the horizontal distance from the axis to the point of intersection (0.7m). The angle is then the arcsin of the opposite side (0.7m) over the hypotenuse (√[(0.7m)2+(0.98m)2]) in radians. (Remember that inverse sin is used to calculate angles). Let's call this angle θ1.
Using Snell's law (n1sinθ1 = n2sinθ2), where the indices of refraction n1 and n2 are for brine (1.52) and oil (1.40) respectively, we can figure out θ2, the angle that the ray makes with the normal in the oil part. The key detail is that angles are always measured from the normal line, which is perpendicular to the interfaces.
Finally, similar calculations will grant what is to be found - the angle the ray makes with the vertical axis in the air (with the index of refraction n=1). This is θ3 calculated from the equation n2sinθ2 = n3sinθ3, where θ2 is the angle that we just found and n2 is the index of refraction for oil. You need to remember that Snell's law applies for any two mediums with a clear border in between through which the light passes.
Finally, the desired angle between the ray's line in the air and the vertical axis will be 90° - θ3 (since θ3 was an angle with the normal line, which is defined as perpendicular to the axis).
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There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot flying low and slow drops a weight; it takes 2.0 s to hit the ground, during which it travels a horizontal distance of 100 m. Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground? How far does it travel before it lands?
Answer:
2.0 s, 200 m
Explanation:
Time to hit the ground depends only on height. Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.
Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance). So it will travel a horizontal distance of 200 m.
Answer:
1) 2 seconds
2) 200 m
Explanation:
1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]
Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]
Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]
s = speed
t = time
h = height
Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,
Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]
i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s
Time taken to land = 2.0 s
2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m
Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]
Therefore, since distance traveled is directly proportional to speed,
Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]
i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]
[tex]2S_{1}[/tex] = 2 x 100 m = 200 m
Distance traveled = 200 m
A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington, directly north to Victoria, British Columbia, a distance of 50 km. An ocean current flows through the Strait of Juan de Fuca from west to east at 3.0 km/h. In what direction (relative to north) should she swim to make the crossing along a straight line between the two cities?
Let [tex]\theta[/tex] be the direction the swimmer must swim relative to east. Then her velocity relative to the water is
[tex]\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]
The current has velocity vector (relative to the Earth)
[tex]\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath[/tex]
The swimmer's resultant velocity (her velocity relative to the Earth) is then
[tex]\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}[/tex]
[tex]\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath[/tex]
We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:
[tex]\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ[/tex]
which is approximately 41º west of north.
The swimmer should swim approximately 36.87 degrees in the north-west direction to counteract the eastward ocean current and reach Victoria, British Columbia directly from Port Angeles, Washington. This is derived by considering the velocity vectors of the swimmer and the current.
Explanation:This question is about the interaction of velocity vectors which is a concept from physics. The swimmer's speed and the ocean current each represent a vector with both a magnitude (speed) and direction. We must factor in these two vectors to work out where the swimmer should aim to swim.
Assuming the north direction as +y axis and east as +x axis, the swimmer intends to swim north with velocity 4.0 km/h and the water's current is moving towards east with velocity 3.0 km/h. In order for the swimmer to move directly north, the swimmer needs to swim against the current, which means westwards. By utilizing the Pythagorean Theorem and trigonometric principles, we can calculate the angle of her direction which is arctan(3/4) in the north-west direction or around 36.87 degrees from north.
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For general projectile motion, when the projectile is at the highest point of its trajectory, its acceleration is zero? A)The horizontal and vertical components of its velocity are zero. B) Its velocity is perpendicular to the acceleration. C)The horizontal component of its velocity is zero. D)Its velocity and acceleration are both zero.
Answer:
B) Its velocity is perpendicular to the acceleration.
Explanation:
For general projectile motion, the horizontal acceleration is 0 and the vertical acceleration is -g. This is true for all points on the trajectory.
At the highest point, the vertical velocity is 0. So you have only a horizontal velocity as well as a vertical acceleration. So the two are perpendicular.
An object moving in a straight line changes its velocity uniformly from 2m/s to 4 m/s over a distance of 12 m. What was its acceleration? (A) 0.5 m/s^2 (B) 1 m/s^2 (C) 2 m/s^2 (D) 3 m/s^2
Answer:A
Explanation:
Initial velocity, u = 2m/s
Final velocity, v = 4m/s
Distance covered, s = 12m
Acceleration, a = ?
Using
v² = u² + 2as
2as = v² - u²
a = v²-u²/2s
a = 4²-2²/2 x 12
a = 16-4/24
a = 12/24
a = 0.5m/s²