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bearhunter [10]
3 years ago
8

A 600-g block is dropped onto a relaxed vertical spring that has a spring constant k =190.0 N/m. The block becomes attached to t

he spring and compresses the spring 37.5 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by the gravitational force on it?
Physics
1 answer:
charle [14.2K]3 years ago
5 0

Answer:

Work done will be 2.205 j

Explanation:

We have given that the spring is compressed b 37.5 cm

So d = 0.375 m

Mass of the block m = 600 gram = 0.6 kg

Acceleration due to gravity g=9.8m/sec^2

Gravitational force on the block F=mg=0.6\times 9.8=5.88N

Now we know that work done is give by W=Fd=5.88\times 0.375=2.205J

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Sedimentary rocks also known as clastic sedimentary rock
3 0
3 years ago
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Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
4 years ago
4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg} (6)

Finally:

V_{f}=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

7 0
4 years ago
A football coach walks 24 meters eastward, then 12 meters
elena-14-01-66 [18.8K]

Answer:

Distance magnitude = 24 meters

The direction of the displacement = Eastward

Explanation:

The distance walked by the football coach eastward = 24 meters

The distance walked by the football coach westward = 12 meters

Then the distance walked by the football coach eastward = 36 meters

Finally, the distance walked by the football coach westward = 22 meters

The total distance walked by the coach eastward  = 24 + 36 = 60 meters

The total distance walked by the coach westward  = 12 + 22 = 34 meters

Taking x-direction as positive, we have;

The magnitude of the distance = 60 - 34 = 24 meters,

The direction of the coach = Eastward.

4 0
3 years ago
Which conclusions about alcohol and teen driving are supported by the data provided? Check all that apply. Rates of drinking and
Kisachek [45]

Answer: • Fewer teens admitted to drinking and driving in 2013 than in 2004.

• In 2013, about one in five teens admitted to riding with a driver who had been drinking.

Explanation:

The conclusions about alcohol and teen driving that are supported by the data provided will be:

• Fewer teens admitted to drinking and driving in 2013 than in 2004.

• In 2013, about one in five teens admitted to riding with a driver who had been drinking.

Other options are incorrect as the conclusion wasn't gotten from the data that was orovufed

8 0
3 years ago
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