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mestny [16]
3 years ago
14

How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?

Physics
1 answer:
ddd [48]3 years ago
6 0

Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

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avanturin [10]

it affects your body by building/straining your muscles so you can be stronger & more athletic

6 0
3 years ago
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QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for
Dominik [7]

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

  • 40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

  • v = v₀ + at

Substitute the known values into the equation.

  • 11.1111 = 0 + (4.5)t
  • 11.1111 = 4.5t
  • t = 2.469133333

The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

5 0
3 years ago
Suppose that the half-life of an element is 1000 years. How many half-lives will it take before one-eighth of the original sampl
34kurt

Answer:

3

Explanation:

The half-life of a radioactive isotope is the time it takes for the mass of the sample to halve.

This can be rewritten as follows:

m(t) = m_0 (\frac{1}{2})^n

where

m(t) is the mass of the sample at time t

m0 is the original mass of the sample

n is the number of half-lives that passed

We see that if we take n=3, the amount of original sample left is

m(t) = m_0 (\frac{1}{2})^3 = m_0 (\frac{1}{8})

So 3 (3 half-lives) is the correct answer.

3 0
3 years ago
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Stephanie is riding her bicycle to the park. The chemical energy used to pedal the bicycle is transformed into __________.
Jet001 [13]

Answer:I'm gonna say mechanical or kinetic depending on how you look at it.

Explanation:

4 0
2 years ago
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A 1425 kg truck driving at 13.0 m/s collides elastically with a stationary 1175 kg car. If the car is traveling 14.25 m/s just a
TEA [102]

Answer:

1.25 m/s

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(1425*13)+(1175*0)=(1425*v1f)+(1175*14.25)

18525+0=1425(v1f)+16743.75

1781.25=1425(v1f)

v1f=1.25 m/s

4 0
2 years ago
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