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trapecia [35]
2 years ago
14

A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding

without any friction on a smooth surface. a. Find the momentum of the box with respect to the floor. b. Find the momentum of the box with respect to the floor after she puts the box down on the frictionless skating surface.
Physics
1 answer:
Yuki888 [10]2 years ago
8 0

For the part a) we need only the momentum of the box and we have the data to find it.

Momentum is given by,

p=mv

where clearly, p is the momentum, m the mass of the box and v is the velocity.

Substituting,

p=(5kg)(5m/s)\\p=25kg.m/s

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

p=(5kg)(5m/s)\\p=25kg.m/s

<em>It is the same that part a)</em>

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A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component
Snezhnost [94]

Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

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We say

/r/= √x²+y²

So

30²= √18² + y²

y= √576

Y= 24m

7 0
2 years ago
What are the strengths and weaknesses of the four methods of waste management?<br>​
Nitella [24]

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* It creates employment

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* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

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4 0
3 years ago
Temperature Dependence of the pH of pure Water
Mamont248 [21]

Answer:

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8 0
3 years ago
On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs
user100 [1]

Answer:

A) \Delta P =  14512.5 Pa = 14.512 kPa

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as v_2 = 150 m/s

since inside air is atmospheric , v_1 = 0 m/s

a) By using bernoulli equation between outside and inside of flight

P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh

\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2

\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]

\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]

\Delta P =  14512.5 Pa = 14.512 kPa

b) force exerted on window

Area of window  = 25\times 45 = 1125 cm^2 = 0.1125 m^2

We know that force is given as

F = P\times A

F = 14512.5 \times 0.1125

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5 0
3 years ago
Find the final velocity of a car that accelerates at +2 m/s2 for +4m from an
Stolb23 [73]

Answer:

Explanation:

according to third equation of motion

2as=vf²-vi²

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vf=√2as+vi²

vf=√2as+vi

vf=√2*2*4+3

vf=√16+3

vf=4+3=7

so final velocity is 7 m/s

5 0
3 years ago
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