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Daniel [21]
3 years ago
10

A 250 kg car has 6875 kg•m/s of momentum. What is it’s velocity?

Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

v = 27.5 m/s

Explanation:

p = m × v

6,875 = 250 × v

250v = 6,875

v = 6,875/250

v = 27.5 m/s

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What is the power of 10 when 0.00503 is written in scientific notation?
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Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f
stealth61 [152]

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put  12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

4 0
3 years ago
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a
lubasha [3.4K]
The angular speed can be solve using the formula:
w = v / r
where w is the angular speed
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w = ( 5 m / s ) / ( 5 cm ) ( 1 m / 100 cm )
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8 0
3 years ago
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While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba
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A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca
sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09  m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

d  = ut + \frac{1}{2}at^2

u = 0 starting from rest

d = \frac{1}{2}at^2

t^2 = \frac{2d}{a}

for bike

d+25 = 0 + \frac{1}{2}*4.40t^2

t^2= \frac{d+25}{2.20}

equating time of both

\frac{2d}{a} = \frac{d+25}{2.20}

solving for d we get

d = 132 m

therefore t is= \sqrt{\frac{2d}{a}}

t =  \sqrt{\frac{2*132}{3.70}}

t = 8.45 sec

each travelled in time 8.45 sec as

for car

d = \frac{1}{2}*3.70 *8.45^2

d = 132.09  m

fro bike

d = \frac{1}{2}*4.40 *8.45^2

d = 157.08 m

7 0
3 years ago
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