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My name is Ann [436]
3 years ago
9

A 1.0kg cart moving right at 5.0 m/s on a frictionless track collides and

Physics
1 answer:
kakasveta [241]3 years ago
6 0
Let’s use the *queue dramatic voice* LAW OF THE CONSERVATION OF MOMENTUM!
m1v1i + m2v2i = m1vf + m2vf

m1v1i - m1vf = m2vf - m2v2i

m1(v1i - vf) = m2(vf - v2i)

m2 = [m1(v1i - Vf)] / (vf - v2i)

m2 = [(1)(5 - -1)] / (-1 - -2)

m2 = 6 / 1

m2 = 6 kg
And that’s your final answer! Please press “Thanks”!
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What are the units for volume and/or mass? Please answer and explain clearly thanks!
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The units for mass are grams (g) and kilograms (kg)
the units for volume are millilitres (ml) and litres (l)
6 0
3 years ago
At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh
Anna35 [415]

We have the equation for electric field E = kQ/d^{2}

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to d^{2}

So, \frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}

E_{1} = 485 N/C

d_{1} = 0.208 cm

d_{2} = 0.620 cm

E_{2} = ?

\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}

E_{2} = \frac{485*0.208^{2}}{0.628^{2}}

E_{2} = 53.20 N/C

7 0
3 years ago
Read 2 more answers
Which change would cause the largest increase in a mountain climber’s gravitational potential energy?
3241004551 [841]
<span>carrying twice the weight and climbing twice as high</span>
5 0
3 years ago
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A car with a mass of 1,500 kg is accelerating at a rate of
AURORKA [14]

Answer:

The net force acting on the car is

3

×

10

3

Newtons.

Hope this helps you

Explanation:

Force is defined as the product of the mass of the body and its aaceleration,

⇒

F

=

m

a

Substituting the above given values we get,

F

=

(

1500

k

g

)

(

2.0

m

/

s

2

)

=

3000

N

=

3

×

10

3

N

.

4 0
2 years ago
A stone is thrown straight downward with a speed of 20 m/s from the top of a tall building. If the stone strikes the ground 3.0
Lady_Fox [76]
Do you remember this formula for the distance traveled while accelerated ?

<u>Distance = (initial speed) x (t)  plus  (1/2) x (acceleration) x (t²)</u>

I think this is exactly what we need for this problem.

initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds

Distance down = (20) x (3)  plus  (1/2) x (9.81) x (3)²

Distance = (60)  plus  (4.905) x (9)

Distance = (60)  plus  (44.145)  =  104.145 meters

Choice  <em>D)</em>  is the closest one.
6 0
3 years ago
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