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AlladinOne [14]
3 years ago
14

Which Newton’s law is shown below? First Second Third Pls answer

Physics
1 answer:
Temka [501]3 years ago
5 0

Answer:

This is Newton's first law.

Explanation:

An object in motion will stay in motion and an object at rest wI'll stay at rest unless acted upon by an unbalanced force.

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A mechanic pushes a 2.60 ✕ 103-kg car from rest to a speed of v, doing 5,430 J of work in the process. During this time, the car
Oduvanchick [21]

Part A. We are given that a car is pushed from rest to a final speed doing 5430 Joules of work. -

To determine the final velocity we will use the work and energy theorem which states that the work done is equal to the change in kinetic energy:

W=K_f-K_0

Since the car starts from rest this means that the initial kinetic energy is zero:

W=K_f

The kinetic energy is given by:

K=\frac{1}{2}mv^2

Substituting in the formula we get:

W=\frac{1}{2}mv_f^2

Now, we solve for the final velocity. First, we multiply both sides by 2 and divide both sides by the mass:

\frac{2W}{m}=v_f^2

Now, we take the square root to both sides:

\sqrt{\frac{2W}{m}}=v_f

Now, we plug in the values:

\sqrt{\frac{2(5430J)}{2.6\times10^3kg}}=v_f

Solving the operations:

2.04\frac{m}{s}=v_f

Therefore, the final velocity is 2.04 m/s.

Part B. Now, we use the following formula for work:

W=Fd

Where "F" is the force and "d" is the distance.

Now, we set this equation equal to the work done by the force:

Fd=5430J

Now, we divide both sides by the distance "d":

F=\frac{5430J}{29m}=187.2N

Therefore, the force exerted is 187.2 Newton.

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Information about climate change?​
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True or false: seismographs from three locations are studied to calculate the epicenter of am earthquake
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3 years ago
(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new
Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation  

m \frac{d^{2}x}{dt^{2}} +kx=0

Definition of parameters  

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

x(t)=c1cos(ωt)+c2Sin(ωt)

x(t)=c1cos(5.929t)+c2Sin(5.929t)

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is  

x(t)=0.337sin((5.929t)

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3 years ago
How to get displacement
andrew-mc [135]
Displacement is how much of a liquid (typically water for simplicity in the metric system) is pushed aside when another object is completely submerged. For example, when a 100mL of water has a block placed into it, and rises to 150mL, the block has displaced the water.
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4 years ago
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