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Virty [35]
3 years ago
9

Initially stationary, a train has a constant acceleration of 0.8 m/s2. (a) What is its speed after 27 s? m/s (b) What is the tot

al time required for the train to reach a speed of 41 m/s?
Physics
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

(a) v1 = 21.6 m/s

(b) t = 51.25 s

Explanation:

Use kinematics equation

v1 = v0 + at

Given

v0 = 0 = initial velocity

a = 0.8 m/s^2 = acceleration

(a) t = 27 seconds

v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s

(b) v1 = 41 m/s

v1 = v0 + at

solve for t

t = (v1-v0)/a = (41-0)/0.8 = 51.25 s

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The density of the material would be 4.1 g/cm³.

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D=m÷v

D=45 g÷11 cm³

D=4.1 g/cm³

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Explain how energy balance sets planetary temperature? Imagine a planet colder than expected for energy balance and explain why
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The planetary temperature energy balance is obtained by radiating back the absorbed radiation energy from outer-space, by the planet and thus acquiring thermal equilibrium.

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6 0
1 year ago
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe
Salsk061 [2.6K]
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

5 0
2 years ago
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otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

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Formula used to calculate period is as follows.

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where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

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