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3 years ago

9

How do motion and Newton's laws apply to your everyday life? (all the laws)

1 answer:

Nataly [62]3 years ago

4 0

For me personally, Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air.

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Find the mass of a 52.2N bucket.

Goshia [24]

Answer:

m = 5.22 kg

Explanation:

The force acting on the bucket is 52.2 N.

We need to find the mass of the bucket.

The force acting on the bucket is given by :

F = mg

g is acceleration due to gravity

m is mass

$m=\dfrac{F}{g}\\\\m=\dfrac{52.2}{10}\\\\=5.22\ kg$

So, the mass of the bucket is 5.22 kg.

3 0

3 years ago

I need help quick!! Urgent!

faltersainse [42]

C.

Explanation: Because mirrors are lighter, and they are easier than lenses to make perfectly smooth.

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3 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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2 years ago

A body of mass 500kg moving at a speed of 10m/s reaches the speed of 50m/s in 20s.The force exerted is

Leona [35]

Answer:

answer is 1000 N

formula used-

<em><u>F= m x (v-u/t)</u></em>

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6 0

3 years ago

A load of 45N attached to a spring that is hanging vertically stretches the spring 0.14m. What is the spring constant?

Vesna [10]

6.3 That Would be the I answer I think but Check on Google For the formula

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3 years ago