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2 years ago

A ball starts from rest and accelerates at a constant rate of 1.0m/s to a final velocity of

Physics

1 answer:

den301095 [7]2 years ago

8 0

Answer:

Time taken to reach final velocity = 5.5 second

Explanation:

Given:

Initial velocity (Starting from rest)(u) = 0 m/s

Acceleration of ball (a) = 1 m/s²

Final velocity (v) = 5.5 m/s

Find:

Time taken to reach final velocity

Computation:

Using first equation of motion;

v = u + at

where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

5.5 = 0 + (1)(t)

5.5 = t

Time taken to reach final velocity = 5.5 second

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What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the mag

zalisa [80]

Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

$N\phi = BAN sin \theta$

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

$\theta$ is the angle between the direction of the field and the normal to the coil

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

$\theta=90^{\circ}$

Therefore, the magnetic flux linkage is

$N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb$

7 0

3 years ago

What is the angular displacement of the wheel between t = 5 s and t = 15 s?

Mkey [24]

The question seems incomplete. The complete text is:

a)What is the angular displacement of the wheel between t = 5 s and t = 15 s

b)What is the angular velocity of the wheel at 15 s

And it refers to the attached figure.

a) 25 rad

The graph shown represent the angular position of the wheel at different times.

Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.

At $t_1 = 5 s$ , the angular position from the graph is $\theta_1 = 100 rad$

At $t_2 = 15 s$ , the angular position from the graph is $\theta_2 = 125 rad$

Therefore, the angular displacement is

$\Delta \theta= \theta_2 - \theta_1 = 125-100 = 25 rad$

2) -5.0 rad/s

For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.

Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.

By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:

$\frac{\Delta \theta}{\Delta t}=\frac{100 rad - 150 rad}{20 s - 10 s}=-5.0 rad/s$

This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s

$\omega = -5.0 rad/s$

5 0

3 years ago

A 0.850 kg mass is placed on a

densk [106]

Answer:

Picture

Explanation:

I am trying to solve this Q

but not sure if this a correct answer

or not.

Hope this helps.

5 0

3 years ago

If a train is going 60 m/s hits the brakes, and it takes the train 1 minute 25 seconds to stop, what is the train’s acceleration

Cloud [144]

Hey there, the answer is .............................. About 0.7 m/sec^2

Acceleration is the change in speed / time 

Change in speed is 60 m/sec 

Time is 1 minute 25 second. Convert that to seconds. 

Divide the change in speed by the time in seconds.

About 0.7 m/sec^2

So the acceleration is - 60 / 85 = - 0.71 m/s^2

HOPE I HELPED!!!!!!!!!!

8 0

3 years ago

A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0

3 years ago