Use a(t) =−32 feet per second squared as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. for how many seconds will the ball be going upward?

Answers

Answer 1

Answer: Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by
$v(t)= v_0 - a t$
where we took upward as positive direction, and where $v_0$ is the initial velocity, a the acceleration and t the time.

The instant at which $v(t)=0$ is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:
$0=v_0 - at$
By plugging numbers into the equation, we find
$t= (v_0)/(a)= (56 ft/s)/(32 ft/s^2)=1.75 s$

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An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

Part 1) A cop car traveling at 25 m/s has a siren producing a frequency of 700 Hz. A felon jumps on his motorcycle and speed off in the opposite direction of 15 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 2) The cop does a U-turn and speeds towards the felon at 30 m/s, while the felon speeds up to 20 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?
Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

Answers

1) 621.8 Hz

2) 719.3 Hz

3) 700 Hz

Explanation:

The Doppler effect occurs when there is a source of a wave in relative motion with respect to an observer.

When this happens, the frequency of the wave appears shifted to the observer, according to the equation:

$f'=(v\pm v_o)/(v \pm v_s)f$

where

f is the real frequency of the sound

f' is the apparent frequency of the sound

v is the speed of the sound wave

$v_o$ is the velocity of the observer, which is negative if the observer is moving away from the source, positive if the observer is moving towards the source

$v_s$ is the velocity of the source, which is negative if the source is moving towards the observer, positive if the source is moving away

In this problem we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=-25 m/s$ is the velocity of the car with the siren

$v_o = +15 m/s$ is the velocity of the felon (he's moving away from the siren)

So, the frequency heard by the felon is

$f=(343-25)/(343+15)(700)=621.8 Hz$

In this case, the cop does a U-turn and speeds towards the felon at 30 m/s.

This means that now the siren is moving towards the observer (so, $v_s$ becomes positive), while the sign of $v_o$ still remains positive.

So we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +20 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+20)(700)=719.3 Hz$

In this case, the felon speeds up to 30 m/s.

This means that now the felon and the siren are moving with the same relative velocity: so, it's like they are not moving relative to each other, so the frequency will not change.

In fact we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +30 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+30)(700)=700 Hz$

So, the frequency will not change.

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

Answers

Answer:

360 Nm

Explanation:

Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).

From the question,

The expression of torque is given as

τ = F×d......................... Equation 1

Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.

Given: F = 40 N, d = 9 m

Substitute into equation 1

τ = 40(9)

τ = 360 Nm.

Answer:

360Nm

Explanation:

Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;

τ = r F sin θ

Where;

r = distance from the pivot point to the point where the force is applied

F = magnitude of the force applied

θ = the angle between the force and the vector directed from the point of application to the pivot point.

From the question;

r = 9m

F = 40N

θ = 90° (since the force is applied perpendicular to the end of the bar)

Substitute these values into equation (i) as follows;

τ = 9 x 40 sin 90°

τ = 360Nm

Therefore the torque is 360Nm

A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushedforward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?

Answers

The answer to your question is 55

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $(1)/(2) kx^(2) +PE1$