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shusha [124]
3 years ago
13

A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t

he hit, the puck has a speed of 40 m/s. The puck comes to rest after going a distance of 30 m. Find the magnitude of the net friction force.

Physics
2 answers:
kirill [66]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

JulsSmile [24]3 years ago
3 0

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

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A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
1 point
WINSTONCH [101]

Answer:

 Mg- 27   means isotope with 12 protons and 15 neutrons.

Also 27 is mass number which express sum of protons and neutrons.

In nucleus one neutrn decays to electron and proton. Mass number remain same but Al-27 nucleus contain 13 protons and 14 neutrons. Electron is ejected out from nucleus.

7 0
3 years ago
During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros
dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

I=\frac{q}{t}

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

I=2.50\cdot 10^4 A is the current

t=40.0 \mu s = 40.0\cdot 10^{-6} s is the time interval

Solving the formula for q, we find the amount of charge transferred:

q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C

6 0
2 years ago
If a proton were released in the electric field above, what direction would it move?
Afina-wow [57]

Answer:

d because the proton would move towards the negative plate

Explanation:

5 0
3 years ago
How will the motion of a falling whirligig compare to that of a falling paper ball?
kicyunya [14]

The motion of a falling whirligig is different to that of a falling paper ball due to spinning.

<h3>Type of motion performed by whirligig and falling paper ball </h3>

The motion of a falling whirligig is different from the motion of a falling paper ball because the paper ball falls on the ground without spinning while on the other hand, the whirligig falls on the ground along with spinning.

The falling whirligig performs two motion i.e. one is falling on the ground and the other is spinning during motion whereas paper ball performs one motion i.e. motion in the air towards the ground so we can conclude that the motion of a falling whirligig is different than of a falling paper ball.

Learn more about motion here: brainly.com/question/453639

4 0
2 years ago
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