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Ann [662]
2 years ago
5

Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap

plied. If the duration for which force acts is extended to 5 s, what will be the velocity of the object then?​
Physics
1 answer:
Alenkasestr [34]2 years ago
7 0

Answer:

Force=10N

Velocity=13m/s

Explanation:

Given,

Mass = 5kg

initial velocity=3m/s

Final velocity=7m/s

Time=2s

Now acceleration= v-u/t

=7-3/2

=4/2

=2m/s²

So, acceleration=2m/s²

Now,

F=ma

=5*2

=10N

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A block of mass mm starts from rest and slides down from the top of a wedge of height hh and length dd. The surface of the wedge
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Answer:

v=\sqrt{ 2 (g sin\theta- f/m) d}

Explanation:

Given that

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The force due to gravity along inline plane

Fg= mg sinθ

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mg sinθ  - f = m a

a= (g sinθ  - f/m)

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2 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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2 years ago
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