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Dmitry_Shevchenko [17]
3 years ago
10

2. A 75 kg runner accelerates from 0.00 m/s to 10.0 m/s in 1.5 seconds.

Physics
1 answer:
raketka [301]3 years ago
3 0
A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2

A.b.) sorry ! I don’t no how to do this question
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A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
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If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
3 years ago
A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc
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Answer:

a= - 6.667 m/s²

Explanation:

Given that

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Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

6 0
3 years ago
A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of
kirza4 [7]
The answer would be 9.8 meters/ sec^2.

7 0
3 years ago
Read 2 more answers
A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is
Reil [10]

We have the equation of motion v^2=u^2+2as, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

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Substituting 40^2=0^2+2*a*2\\ \\ a=400m/s^2

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4 0
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Answer:

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Explanation:

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for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

8 0
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