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3 years ago

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(I) Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.0 min to tur

n through 23 complete revolutions before reaching its final speed. (a) What was its angular acceleration (assumed constant), and (b) what was its final angular speed in rpm?

1 answer:

Eddi Din [679]3 years ago

3 0

Answer:

The angular acceleration is $0.08rad/s^{2}$ final angular speed is $46rpm$

Explanation:

Here we know that

$\theta= w_{o}t+\frac{1}{2}\alpha t^{2}$ .............. 1

$\theta = 23\times 2\pi rad$

$w_{0}=0\\t=1min =60s$

Upon substituting these values in equation 1 we get $\alpha =0.08rad/s^{2}$ .

For calculating angular velocity we know that

$w=w_{o} + \alpha t$ .

$w=\frac{46\pi}{30}rad/s = \frac{23\pi\times60}{15\times 2 \pi} rpm = 46rpm$

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a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in

zhannawk [14.2K]

Answer:

a. F = 245 Newton.

b. Workdone = 392 Joules.

c. Power = 196 Watts

Explanation:

Given the following data;

Mass = 25kg

Distance = 1.6m

Time = 2secs

a. To find the force needed to lift the mass (in N );

Force = mass * acceleration

We know that acceleration due to gravity is equal to 9.8

F = 25*9.8

F = 245N

b. To find the work done by the student (in J);

Workdone = force * distance

Workdone = 245 * 1.6

Workdone = 392 Joules.

c. To find the power exerted by the student (in W);

Power = workdone/time

Power = 392/2

Power = 196 Watts.

5 0

2 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago

Look at the Grand Canyon cross section. What is the 4th event to take place in the Grand Canyon's geologic history?

mart [117]

Answer:

bass limestone

Explanation:

if I'm right bass limestone is the fourth one because 3 are under it

8 0

2 years ago

Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an

trapecia [35]

Answer: A. Her speed is 4.4 m/s, and her velocity is 0 m/s.

Explanation: i took the test on edgenuity

6 0

3 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

$a=-1 m/s^2$ is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

$T=-\frac{u}{a}=-\frac{10}{-1}=10 s$

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

$a=-1 m/s^2$ is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

$s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m$

7 0

3 years ago