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3 years ago

a bus accelerates at -0.5 m/s^2 for 12.5 seconds if the bus had an initial speed of 31 /s what is its new speed

Physics

2 answers:

Arlecino [84]3 years ago

8 0

Answer:

as the other person said, the answer is 24.75 m/s

Explanation:

aleksandr82 [10.1K]3 years ago

3 0

Vf=v0+at so 31m/s-.5m/s^2*12.5= 24.75m/s is the new speed. the negative sign is showing a deceleration.

Hope this helps! Please rate if you like my answer! Thank you!!

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Work is the transfer of _______ that occurs when a force makes an object move.

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3 years ago

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid

Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

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3 years ago

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3 years ago

What is the pressure exerted on three floor by a 10.000N crate if the bottom of the crate has the dimensions of 1/2m by 4m

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Answer:

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Explanation:

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6 0

3 years ago

The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb

topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

$\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\$

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

$I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2$

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

$F = \frac{mv^2}{r}$

The relation between linear velocity and the angular velocity is

$v = \omega r$

So,

$F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}$

As can be seen, the maximum velocity for the projectile without breaking the cable is $\sqrt{1500r}$ , where r is the length of the cable.

6 0

3 years ago