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umka2103 [35]
3 years ago
14

Find the Acceleration of a car with the mass of 1200kg and a force of 11x10^3 N

Physics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

<h2>9.17 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{11 \times  {10}^{3} }{1200}  \\  = 9.16666...

We have the final answer as

<h3>9.17 m/s²</h3>

Hope this helps you

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A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d
Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

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6 0
3 years ago
Where is the value of kinetic energy equal to zero
bearhunter [10]

Answer:

if the object is not in motion

Explanation:

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Read 2 more answers
n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic
Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m  + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m  + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0
3 years ago
Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm
Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

q_{1} = charge on one sphere

q_{2} = charge on other sphere

q = Total charge on the two spheres = 40 μC

q_{1}+ q_{2} = q

q_{1}+ q_{2} = 40 x 10⁻⁶

q_{1} = (40 x 10⁻⁶) - q_{2}                                   eq-1

r = distance between the two spheres = 50 cm = 0.50 m

F = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

F = \frac{k q_{1} q_{2}}{r^{2}}

2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}

q_{2} = 1.44 x 10⁻⁶ C

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3 years ago
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