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yanalaym [24]
3 years ago
11

The rate at which velocity changes is called

Physics
1 answer:
grandymaker [24]3 years ago
6 0
The answer is acceleration.

Acceleration is the rate at which velocity changes over time. The relationship between acceleration and velocity is presented in the formula:
a = Δv/Δt
The SI unit of acceleration is m/s² derived as following:
a = Δv/Δt [ = ] (m/s)/s = m/s * s = m/s²
You might be interested in
When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to
Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

<em>It took Kevin 26 minutes to run from markers 1 to 4</em>

<em></em>

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = \frac{2}{13} = 0.154 miles/minute

<em>His average speed from mile markers 1 to 4 is 0.154 miles/minute</em>

<em></em>

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

<em>Kevin must run by average speed 0.1 miles/minute to finish the race</em>

7 0
2 years ago
Read 2 more answers
Did I do these questions correctly?
Elena L [17]
For number 11, you should say that there is more pollution and fossil fuels being burned.
8 0
2 years ago
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
Will give brainliest, Pleaseee help!!!
puteri [66]

Answer:

below

Explanation:

1.1115 im not sure tho

5 0
2 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
3 years ago
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