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Hunter-Best [27]
4 years ago
12

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

Note: 1 kg = 1000 g).
Physics
1 answer:
slavikrds [6]4 years ago
6 0

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is p=7.2\times 10^{-3}\ kg-m/s

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, p=7.2\times 10^{-3}\times 10^3\ g-m/s

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

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Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.
lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0
3 years ago
true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.
nalin [4]
Nope. It's called 'centripetal' acceleration. The force that created it MAY be gravitational, but it doesn't have to be. For things on the surface of the Earth moving in circles, it's never gravity.
5 0
3 years ago
A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00
marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)        (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

3 0
4 years ago
Based on the image below what would an observer on the nighttime side of earth observe?
sertanlavr [38]
I would love to help; where’s the picture?
3 0
3 years ago
A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is
ollegr [7]
<span>F = ma
</span>Ff = μ*Fn
<span>Fn = Fw
</span>Fw = mg 

<span>So we have: </span>

<span>Ff = μmg </span>

<span>And </span>

<span>Ff = ma </span>

<span>So... </span>

<span>μmg = ma </span><span> </span>

<span>μg = a </span>

<span>And we can solve for the acceleration: </span>

<span>(0.15)(9.81 m/s²) = a </span>

<span>a = 1.47 m/s² </span>
8 0
3 years ago
Read 2 more answers
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