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Marysya12 [62]
3 years ago
10

A vehicle has an initial velocity of 30 meters per second. 30 seconds later it is traveling at a velocity of 60 meters per secon

d. what is the acceleration of the vehicle?
a) 10 meters/second/second in the opposite direction of travel
b) 10 meters/second/second in the same direction of travel
c) -1 meter/second/second in the same direction of travel
d) 1 meter/second/second in the same direction of travel
Physics
2 answers:
Maslowich3 years ago
5 0
I would say the answer is b hope I helped
vekshin13 years ago
3 0
D is your answer if im right
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Doubling an object’s height will have what effect on its potential energy due to gravity?
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Potential energy due to gravity = Ep = mgh [symbols have their usual meaning ]
Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.
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A skateboarder is moving at 2.00m/s and increases his velcity at a rate of 1.6m/s/s for 6.0 seconds. What is the displacement of
MrRissso [65]

Answer:

S = 40.8m

Explanation:

<u>Given the following data;</u>

Initial velocity, u = 2m/s

Acceleration, a = 1.6m/s²

Time, t = 6secs

Required to find the displacement

Displacement, S = ?

The displacement of an object is given by the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Where;

  • S represents the displacement measured in meters.
  • u represents the initial velocity measured in meters per seconds.
  • t represents the time measured in seconds.
  • a represents acceleration measured in meters per seconds square.

<em>Substituting into the equation, we have;</em>

S = 2*6  + \frac {1}{2}*(1.6)*6^{2}

S = 12 + 0.8*36

S = 12 + 28.8

S = 40.8m

<em>Therefore, the displacement of the skateboarder during this game is 40.8 meters. </em>

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3 years ago
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WINSTONCH [101]
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3 years ago
What is kinematics???​
dalvyx [7]

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An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C.225 ∘C. During this process, 346 J of work was done by the system
Ne4ueva [31]

Answer:

402 J/(kg ⁰C)

Explanation:

m = mass of the sample = 80.0 g = 0.080 kg

T₀ = Initial temperature of the sample = 25 ⁰C

T = Final temperature of the sample = 225 ⁰C

W = Amount of work done by the system = 346 J

U = Increase in internal energy = 6085 J

Q = Amount of heat given to the gas

Amount of heat is given as

Q = U + W

Q = 6085 + 346

Q = 6431 J

Heat received by sample is given as

Q = m c (T - T₀ )

inserting the values

6431 = (0.080) c (225 - 25)

c = 402 J/(kg ⁰C)

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