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3 years ago

Can anyone tell me the ans of this question also please

Physics

2 answers:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

Hey there!

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

Stopwatch Y recorded 10 seconds longer than stopwatch X.

Hope this helps :)

vladimir2022 [97]3 years ago

3 0

Answer:

Explanation:

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

Stopwatch Y recorded 10 seconds longer than stopwatch X.

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A flat uniform circular disk (radius = 2.30 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the

Vadim26 [7]

The resulting angular speed = 0.6 rad / s.

<u>Explanation:</u>

Here there is no external torque acting on the system thus we can apply the law of conservation of angular momentum

Angular momentum of the man = Iω

Where I = Inertia of the man about the axis of rotation

or I = M r 2

I = 50 * 1.25*1.25 = 78.125

w = Angular velocity of the man, that can be calculated as follows

Tangential velocity of man = v = 2m/s

So time taken to describe this circle is t = (2*pi* r) / v

Now angle described in 1 revolution θ = 2*pi radians

This angle is subtended in time t = (2*pi* r) / v

Thus angular speed = w = θ/t = 2*pi* ( v/ 2π r) = v/r = 2.70 / 1.25 = 2.16 rad/s

So angular momentum of man = Iw = 78.125 * 2.16 = 168.75.

To conserve the angular momentum before and after,

Angular momentum of disk = angular momentum of the man

i.e. Iw of disk = 168.75

disk of I = (disk of M*R^2) / 2

= (1.00 * 102 * 2.30 * 2.30) / 2

= 269.79

Thus 269.79 of disk of w = 168.75

Resulting angular speed of disk = 168.75 / 269.79 = 0.6 ras / s

7 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

The image shows the electric field lines around two charged particles.

DENIUS [597]

It's either 3 or 4 I know this becuase I have read a book about electricity

8 0

3 years ago

Read 2 more answers

A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap

Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = $I$ 'ω'

where

$I$ ' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = $mr'^{2}$

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to $I$ .

From

$I$ 'ω' = $I$ ω

since $I$ is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

5 0

3 years ago

The equation for the conservation of mechanical energy is:

marusya05 [52]

Answer:

The answer is A because the equation is KEi+PEi=KEf+PEf

i means initial (before) and f means final (after)

7 0

2 years ago