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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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8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

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Elena-2011 [213]

By definition we have to:

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Some properties are:

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8 0
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Answer:

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