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Nuetrik [128]
3 years ago
10

assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for

ce
Physics
1 answer:
Vesnalui [34]3 years ago
5 0
The sum of force is 7,680...............
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A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc
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Answer:

Explanation:

From A to B

distance traveled with velocity v_1  in time t_1

\frac{d}{2}=v_1t_1----1

from B to C

distance traveled is 0.5 d with v_2  and v_3  velocity for half-half time

\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2

divide 1 and 2 we get

\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}

\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}

Now average velocity is given by

v_{avg}=\frac{d}{t_1+t_2}

taking t_1  common

v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}

v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}

v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}  

6 0
3 years ago
iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig
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Answer:

453 gm

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2 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

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