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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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Help please I will mark you the brainly!

Fudgin [204]

Explanation:

Coz the position changes with time but the position change is not constant throughout time So, D) does not have constant velocity

3 0

3 years ago

Read 2 more answers

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is