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OleMash [197]
3 years ago
10

When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second

Æ. What is the mass of the book?
Physics
1 answer:
4vir4ik [10]3 years ago
8 0
Force = (mass) x (acceleration)    (Newton's second law of motion)

Divide both sides of the equation by 'acceleration', and you have

Mass = (force) / (acceleration)

Mass = 17 newtons / 3.75 meters per second-sqrd = 4.533 kilograms (rounded)
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Answer:

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An unstable nucleus which has a tendency to spontaneously change its form with the emission of high-energy particles or photons
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A rocket has total mass Mi=360 kg , including Mf=330kg of fuel and oxidizer. In interstellar space, it starts from rest at the p
steposvetlana [31]

let the parameters are:

M=instantaneous mass of the rocket

v=velocity of the rocket

t=time

initial mass=Mi=360 kg

fuel mass=Mf=330 kg

relative speed=Ve=1500 m/s

rate of mass decay=k=2.5 kg/s

writing newton’s second law of motion:

                    M*du/dt=-Ve*dM/dt

                           ==>du=-Ve*dM/M

Integration of both sides,

                       u=-Ve*ln(M)

using the limits with at t=0, u=0 and velocity being v(t),

at t=0, mass=Mi, at any time t, mass=Mi-k*t

                 v(t)-0=-Ve*ln((Mi-k*t)/Mi)

                  v(t)=-Ve*ln(1-(k*t/Mi))

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6 0
1 year ago
What’s the units of specific heat
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6 0
3 years ago
An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at
Deffense [45]

Answer:

v_{o}=-14.60m/s

Explanation:

<u>Kinematics equation for first Object:</u>

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

v_{o}=0m/s       The initial velocity is zero

y_{o}=20m

it reach the water at in instant, t1, y(t)=0:

0=y_{o}-1/2*g*t_{1}^{2}

t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s

<u>Kinematics equation for the second Object:</u>

The initial velocity is zero

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

y_{o}=20m

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}

v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s

The velocity is negative, because the object is thrown downwards.

6 0
3 years ago
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