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djyliett [7]
3 years ago
8

An object of mass 9.00 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless le

vel surface. If the spring constant is 95.0 N/m and the spring is stretched by 10.0 cm , what is the magnitude of the acceleration of the object?
Physics
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

Acceleration of the object will be 1.05m/sec^2

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by F=kx=95\times 0.1=9.5N

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So 9.5=9\times a

a=1.05m/sec^2

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saw5 [17]

Answer:

Parkour is a type of sport or “Hobby”

Explanation:

Parkour is the activity or sport of running through an area, typically in an urban environment.

You can perform parkour by jumping over certain objects, leaping and turning, etc.

Hope this helped!

~Oreo!

4 0
3 years ago
5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?
GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

Learn more:

brainly.com/question/17506203?referrer=searchResults

3 0
2 years ago
A 5kg rock and a 10kg rock are dropped from a height of 10m?
fredd [130]
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5 0
2 years ago
Define pressure in short
Makovka662 [10]

Answer:

hope it helps......

Explanation:

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7 0
3 years ago
A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m

The bullet travels a distance of 276.5 m.


5 0
3 years ago
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