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2 years ago

A car has an acceleration of 4m/s^2 to the east for 5 seconds. What is the change in velocity?

Physics

1 answer:

Natali5045456 [20]2 years ago

3 0

Answer:

a= 4m/s^2

t = 5 secs

v = at

v = 4 × 5

v =20 m/s

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Sveta_85 [38]

Explanation:

the answers are the first 3.

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2 years ago

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f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

vazorg [7]

Answer:

The magnitude of the angular acceleration is $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

The angular speed of CD is $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

time taken to decelerate is $t_{CD} = 2.60\ s$

The final angular speed is $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

$a = \frac{w_f - w_{CD}}{t}$

substituting values

$a = \frac{0 - 52.37}{2.60}$

$a = - 20.14 rad/s^2$

The negative sign show that the CD is decelerating but the magnitude is

$a = 20.14 rad/s^2$

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2 years ago

A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h

katen-ka-za [31]

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

mass of rocket, $m_r=50\ g$

time of observation, $t=2\ s$

mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

velocity of air, $v_a=53\ m.s^{-1}$

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$

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2 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

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2 years ago

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if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t

VikaD [51]

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

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2 years ago