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4 years ago

How deep can an object with 6360N hitting on a sponge get ???

Physics

1 answer:

borishaifa [10]4 years ago

5 0

Answer:

The sponge must go $\dfrac{636}{m}\ \text{meter}$ deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

$h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}$

So, the sponge must go $\dfrac{636}{m}\ \text{meter}$ deep.

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The electric field everywhere on the surface of a thin, spherical shell of radius 0.725 m is of magnitude 899 N/C and points rad

Igoryamba

Answer:

Q = -52.56 nC

Explanation:

The electric field outside a spherical shell is given by

$E = \frac{k*Q}{r^{2} }$

where:

Q is the net charge

r is the radius of the spherical shell and E is the electric field magnitude.

Given:

r = 0.725 m

since the field points radially toward the center of the sphere, E = -899 N/C.

k is a constant with a value = 8.99 x 109 Nm²/C².

∴ $-899 = \frac{8.99*10^{9} *Q}{0.725^{2} }$

$Q = \frac{-899*0.725^{2} }{8.99*10^{9} }=-5.256*10^{-8}$

Q = -52.56 nC

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3 years ago

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4 years ago

Read 2 more answers

Two particles of equal masses (m = 5.5x10-15 kg) are released from rest with a distance between them is equal to 1 m. If particl

Tcecarenko [31]

To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.

By Coulomb's law, electrical energy is defined as

$EE = \frac{kq_1q_2}{d}$

Where,

EE = Electrostatic potential energy

q= charge

d = distance between the charged particles

k = Coulomb's law constant

While kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m= mass

v = velocity

There by conservation of energy we have that

EE= KE

There is not Initial kinetic energy, then

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2$

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2$

$v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}$

$v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}$

Replacing with our values we have,

$v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}$

$v_f = 2.802*10^7m/s$

Therefore the speed of particle B at the instat when the particles are 3m apart is $2.802*10^7m/s$

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4 years ago

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3 years ago

A 0.2-kg hockey puck is sliding along the ice with an initial speed of 12 m/s when a player strikes it with his stick, causing i

iVinArrow [24]

Answer:

The impulse the stick applies to the puck is 7 kg-m/s and it is in opposite direction

Explanation:

Given that,

Mass of the hockey puck, m = 0.2 kg

Initial speed of the hockey, u = 12 m/s

Final speed of the hockey, f = -23 m/s (as it reverse its direction)

We need to find the impulse the stick applies to the puck. It is given by the change in momentum of the object. It is given by :

$J=m(v-u)$

$J=0.2\ kg\times (-23-12)\ m/s$

J = -7 kg-m/s

So, the impulse the stick applies to the puck is 7 kg-m/s and it is in opposite direction. Hence, this is the required solution.

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4 years ago