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3 years ago

Determine the tension in the string that connects M2 and M3.

1 answer:

Ivan3 years ago

4 0

<h3>therefore mass m1=4.8 kg and the tension </h3><h3>in the horizontal spring T2=10N.</h3>

<h2>HOPE IT HELPS YOU </h2>

<h2>PLEASE Mark me as brainliest.</h2>

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What caused the blackout in Canada?

Maru [420]

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.

7 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

Given:

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

Assume:

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

4 years ago

When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

ryzh [129]

Answer:

The pendulum bob swing past the mean position because:

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is $-w^{2} A$ and the the acceleration decreases as $-w^{2} x$ towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0

3 years ago

a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after

STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0

3 years ago

Name a situation when potential energy is useful

adell [148]

Gravitational energy is a form of potential energy because it is dependent on the mass of an object and needs to be calculated for the specific object.

3 0

3 years ago