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laiz [17]
3 years ago
5

A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is

moving at a velocity of –7 m/s. After colliding, the 35-gram ball moves at a velocity of –15 m/s. Assume there is no net force on the system. What is the velocity of the 75-gram ball after the collision? Round to the nearest tenth. –4.2 m/s –0.5 m/s 0.5 m/s 4.2 m/s
Physics
2 answers:
Serhud [2]3 years ago
7 0

Answer:

D. 4.2

Explanation:

On Edgenuity :)

elena-s [515]3 years ago
5 0

Answer:

4.2 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v

315 g m/s − 525 g m/s = -525 g m/s + (75 g) v

315 g m/s = (75 g) v

v = 4.2 m/s

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Answer:

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Explanation:

6 0
3 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

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3 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
2 years ago
A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12
bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

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To find: Coefficient of static friction  

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair  

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The F_n is equal to the weight multiplied by its gravity

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μ_s=F_{max}/mg

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= 0.67

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Answer:

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