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avanturin [10]
3 years ago
11

A motorist driving at 25 meters/second decelerates to

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.

Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,

Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules

Just ignore the negative value for the final result because work is a scalar quantity.
andrew11 [14]3 years ago
3 0

Answer:

C.

3.0 × 10^5 joules

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A(n) 0.576 kg soccer ball approaches a player horizontally with a speed of 20.9 m/s. The player illegally strikes the ball with
Travka [436]

Answer:

17.6kg.m/s in the opposite direction of the initial velocity.

Explanation:

In order to solve this we can use the Impulse-Momentum theorem:

J=\Delta p\\J=m(v_f-v_o)

We will assume the motion of the ball approaching the player as positive, so:

J=0.576kg((-9.73m/s)-20.9m/s)=-17.6kg.m/s

So the impulse is 17.6kg.m/s in the opposite direction of the initial velocity.

4 0
3 years ago
[2.21] Please help me find a) and b)
user100 [1]

Answer:

A. 28.42 m/s

B. 41.21 m.

Explanation:

A. Determination of the initial velocity of the ball:

Time (t) to reach the maximum height = 2.9 s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) =?

Thus, we can obtain the initial velocity of the ball as follow:

v = u + gt

0 = u + (–9.8 × 2.9)

0 = u – 28.42

Collect like terms

u = 0 + 28.42

u = 28.42 m/s

Therefore, the initial velocity of the ball is 28.42 m/s.

B. Determination of the maximum height reached.

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) = 28.42 m/s.

Maximum height (h) =?

Thus, we can obtain the maximum height reached by the ball as follow:

v² = u² + 2gh

0² = 28.42² + (2 × –9.8 × h)

0 = 807.6964 + (–19.6h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = –807.6964 / –19.6

h = 41.21 m

Therefore, the maximum height reached by the ball is 41.21 m

8 0
3 years ago
A sprinter in a 100-m race accelerates uniformly for the first 71 m and then runs with constant velocity. The sprinter’s time fo
adoni [48]

Answer:

The acceleration of the sprinter is 1.4 m/s²

Explanation:

Hi there!

The equation of position of the sprinter is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the sprinter at a time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:

x = 1/2 · a · t²

At t = 9.9 s, x = 71 m

71 m = 1/2 · a · (9.9 s)²

2 · 71 m / (9.9 s)² = a

a = 1.4 m/s²

The acceleration of the sprinter is 1.4 m/s²

3 0
3 years ago
9. Which of the following is NOT a recommendation to help you succeed in this course?
nalin [4]

Answer:

Wait until the end of the semester to complete all work.

Explanation:

this is the obvious answer.

please mark me brainliest

8 0
3 years ago
Read 2 more answers
Three oxygen isotopes
igomit [66]
Stable isotopes, radioisotopes, oxygen

6 0
2 years ago
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