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xeze [42]
2 years ago
6

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe

en the puck and ice is 0.11, how much time does it take the puck to come to rest?
Physics
1 answer:
Salsk061 [2.6K]2 years ago
5 0
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

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A basketball star exerts a force of 3225 n (average value) upon the gym floor in order to accelerate his 76.5-kg body upward. de
ryzh [129]
F = m . g  = 76.5 x 9..8 = 749.7
Net Force = 3225 - 749.7 = 2475.3

F = m.a
2475.3 = 76.5 a

a  = 32.35



V = at + v1
V = at + 0
V = 32.35 x 0.15
V = 4.8525

Hope this helps

8 0
3 years ago
Which statement is true about an object that sinks in water?
Rina8888 [55]
A > Is the correct answer :)
5 0
3 years ago
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How far does a 1.2g bullet with kinetic energy of 1.2 j go in 2 seconds ?<br><br><br>​
almond37 [142]

Answer:

To calculate the energy in joules, simply enter the mass of ammunition (in grams) that you use, and the fps that you've read from your Chrono unit.

5 0
2 years ago
The process by which land remove surface materials is called
Citrus2011 [14]

Answer:

Deflation

Explanation:

The process by which wind removes surface materials is called deflation.

Deflation is a process by which wind erodes the Earth's exterior and the regions which experience severe erosion are called deflation zones.

deflation originates by the erosive force of a wind that removes the loosened area and this process is facilitated by a dry climate and a loss of vegetative cover that helps to lose the sediment.

deflation is common in the arid regions.

3 0
3 years ago
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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the
castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

4 0
3 years ago
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