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Blababa [14]
2 years ago
12

when there is a constant unbalanced force applied in the opposite direction as the motion, speed will __________ and acceleratio

n will __________
Physics
1 answer:
Nutka1998 [239]2 years ago
6 0
brainly.com/question/20066572
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Waves are ?<br> that can travel through matter.
Rom4ik [11]

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

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3 years ago
Waves<br> Please need help fast
mixas84 [53]
I believe its the third answer
6 0
2 years ago
A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it
soldi70 [24.7K]

Answer:

The position function is s_{t}=2t^3+5t^2-5t+9.

Explanation:

Given that,

Acceleration a =12t+10

Initial velocity v_{0} = -5\ cm/s

Initial displacement s_{0}=9\ cm

We know that,

The acceleration is the rate of change of velocity of the particle.

a = \dfrac{dv}{dt}

The velocity is the rate of change of position of the particle

v=\dfrac{dx}{dt}

We need to calculate the the position

The acceleration is

a_{t} = 12t+10

\dfrac{dv}{dt} = 12t+10

a_{t}=dv=(12t+10)dt

On integration both side

\int{dv}=\int{(12t+10)}dt

v_{t}=6t^2+10t+C

At t = 0

v_{0}=0+0+C

C=-5

Now, On integration again both side

v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt

s_{t}=2t^{3}+5t^2-5t+C

At t = 0

s_{0}=0+0+0+C

C=9

s_{t}=2t^3+5t^2-5t+9

Hence, The position function is s_{t}=2t^3+5t^2-5t+9.

7 0
3 years ago
Why is the universe here and why do we exist?
8_murik_8 [283]
Become a physicist and maybe one day you can tell me.
8 0
3 years ago
An astronaut on a strange new planet finds that she can jump up to a height of 72 m before she would fall back down to the surfa
gregori [183]

Answer:

25m/s²

Explanation:

Using one of the equations of motion.

v² = u²+2as where

v is the final velocity of the astronaut

u is his initial velocity

a = -g (the acceleration will be acceleration due to gravity since he is acting under the influence of gravity. The value is negative because the astronaut jumps up to a particular height)

s = H = total height covered

The equation will then become;

v² = u²-2gH

Given

u = 60m/s

v = 0m/s

g = ?

H = 72m

Substituting the given value into the equation;

0² = 60²-2g(72)

0 = 3600-144g

-3600 = -144g

g = -3600/-144

g = 25m/s²

The magnitude of his acceleration due to gravity on the planet is 25m/s²

7 0
3 years ago
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