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seropon [69]
3 years ago
10

I need c only please answer it

Physics
1 answer:
Elodia [21]3 years ago
5 0
It will be moving at high speeds
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If the vertical component of velocity for a projectile is 7.3 meters/second, what is its hang time?
Kamila [148]
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.

So, the hang up time, or the time the object stayed on air is calculated using this equation:

a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
6 0
3 years ago
Which of these substances has the greatest index of refraction?
Inessa05 [86]
D. diamond, is the best option..
7 0
3 years ago
Read 2 more answers
A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

3 0
3 years ago
A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0
3 years ago
A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the
DiKsa [7]

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

4 0
3 years ago
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