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3 years ago

For a projectile with no air resistance, at the peak of its path, _______________.

Physics

1 answer:

3241004551 [841]3 years ago

4 0

Answer: "For a projectile with no air resistance, at the peak of its path, it's velocity is equal to zero"

Explanation:

Suppose that you throw an object up. The initial vertical velocity will be positive, and the acceleration (the gravitational acceleration) will point downwards, meaning that it opposes to the initial direction of the velocity, and that decreases the velocity as the time goes by.

There will be a point where the velocity ( that was positive until now, so until now the height of the object was increasing), is equal to zero, this means that at this moment the object stops moving, and after, because we still have negative acceleration, the velocity will start become negative, and the object will fall down.

Then that point where the velocity was zero is the peak of its path, then we can conclude that:

"For a projectile with no air resistance, at the peak of its path, it's velocity is equal to zero"

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the result is the quantization of __Energy__ of the particle

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4 years ago

Identify the factors that cause melting and thus the formation of magma. Choose one or more: A. decompression B. compression C.

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Answer: C Heat Transfer, E Addition of Volatiles and A Decompression

Explanation:

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3 years ago

A car has a force of 2000N and a mass of 1000kg. What is the acceleration

Anastaziya [24]

Answer:

2m/s/s

Explanation:

The formula goes- F=MA

F-Force M-Mass & A-Acceleration

We need to rearrange this formula to find the acceleration-

A=F/M

All we need to do now is substitute the values in

A=2000N/1000kg

A=2m/s^2

In the given option the last option (2m/s/s) would be the ans, as it's the same as 2m/s^2

So ya, I guess that's all

6 0

2 years ago

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

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3 years ago

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max2010maxim [7]

Answer:

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3 years ago