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mamaluj [8]
3 years ago
14

How do you calculate the change in speed?

Physics
2 answers:
Dennis_Churaev [7]3 years ago
8 0

v = d/t

d = distance

t = time

the change in speed (Δv)

\tt \Delta v=\dfrac{d_2-d_1}{t_2-t_1}

marta [7]3 years ago
3 0
Multiply the acceleration by time to obtain the velocity change: velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed. You can convert units to km/h by multiplying the result by 3.6: 27.8 * 3.6 ≈ 100 km/h .
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A car changes its speed by 2 meters per second each second. What is its acceleration?
KatRina [158]

Answer:

2 m/s²

Explanation:

If changes speed by 2 meters per second each second means:

2 m/s²

Because it changes constantly it veloctity.

Remember the aceleration changes the velocity.

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What is the amplitude of the wave shown in the diagram?
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The amplitude is from the absolute value of the 0 point on the y-axis to the highest(peak) or lowest(troph) point of the wave. In this question, 3cm is the highest and -3cm is the lowest, so the amplitude is 3cm.
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Stick a fork into each end of the root vegetable as shown. The forks should be on the same side of the vegetable, and
zhuklara [117]

Answer:

Yes

Explanation:

There is a position that works better than this and that is switching the sides of the forks.

7 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what
Andrews [41]

Answer:

  v=9.6 km/s

Explanation:

Given that

The mass of the car =  m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system  then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5  v  = 48

v=\dfrac{48}{5}\ km/s

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

3 0
3 years ago
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