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blondinia [14]
3 years ago
9

A spring scale exerts a net force of 8.5 N on an object. What is the object's mass if it has an acceleration of

Physics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

mass of the object is 2.18 kg

Explanation:

Given

Force (F) = 8.5 N = 8.5 kg.m/s^{2}

acceleration (a) = 3.9 m/s^{2}

Mass (m) = ?

We know that the newton's second law of motion gives the relation between mass of ab object. force acted upon and the amount the object is accelerated. It is expressed in the form of an equation:

F = ma

mass, m = F/a

               = \frac{8.5 kgms^{-2} }{3.9 ms^{-2} }

               = 2.18 kg

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Find the hiker’s gravitational potential energy if the cliff is 60m high
Furkat [3]

Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

g=9.81m/s^2

h=60m

So, the potential energy

U=m(9.81m/s^2)(60m)\\\\U=588.6*m

m is the mass of the hiker, wich is not in the description of the problem.

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2 years ago
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
Usimov [2.4K]

Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}

v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

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M/s^2 is the correct answer
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