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3 years ago

Directions: WRITE a summary of the Newton’s Laws of Motion reading.

Physics

1 answer:

djyliett [7]3 years ago

8 0

Answer:

In the Newton's first law of motion, a stationary body or that moving with uniform motion will only chnage its state if acted upon by other forces externally.This kind of resistance is caused by the force of inertia.In his second law of motion,when a body is acted by a force, the force will be equal to the product of mass and acceleration of the body.i.e F=ma. In the third law of motion, it explains that a body will exert an equal and opposite force when interacting with another body.Thus, every action force experiences an equal and opposite reaction force.Here the momentum of the bodies will be conserved.

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A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the averag

andrey2020 [161]

Answer:

The small car and the truck experience the same average force.

Explanation:

Here we need to remember two of Newton's laws.

The second one says that:

F = m*a

force equals mass times acceleration.

And the third one says that;

"If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A"

From the third law, if the car experiences a force F due to the impact with the truck, then the truck experiences the same force F due to the impact.

But this seems odd, because we would expect to see the car being more affected by the impact, right?

Well, this is explained by the second law.

Suppose that the mass of the car is m, and the mass of the truck is M.

such that M > m

Then for the small car we have:

F = m*a

And for the truck:

F = M*a'

Because the force is the same for both of them, we can write:

m*a = M*a'

a = (M/m)*a'

because M > m, then M/m > 1.

This means that the acceleration that the car experiences is larger than the acceleration for the truck, and this is why we would see that the car seems more affected by the impact, regardless of the fact that both vehicles experience the same force in the impact.

6 0

2 years ago

A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket

frutty [35]

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

Net Force = 399532

7 0

3 years ago

A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
$P = i^2R$

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

$P = (2.0^2)(50) = \boxed{\text{ B. }200 W}$

7 0

2 years ago

if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?

laiz [17]

Answer:

The 39.

Explanation:

8 0

2 years ago

Read 2 more answers

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago