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Rina8888 [55]
3 years ago
5

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

Physics
1 answer:
lozanna [386]3 years ago
7 0
You can find

1) time to hit the ground
2) initial velocity
3) speed when it hits the ground

Equations

Vx = Vxo

x = Vx * t

Vy = Vyo + gt

Vyo = 0

Vy = gt

y = yo - Vyo - gt^2 / 2

=> yo - y = gt^2 / 2

1) time to hit the ground

=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

=> t = √1.631 s^2 = 1.28 s

2) initial velocity

Vxo = x / t = 6.5m / 1.28s = 5.08 m/s

3) speed when it  hits the ground

Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s

V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2

=> V = √ (183.56 m^2 / s^2) = 13.55 m/s
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a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the acceleration of the box, we apply Newton's second law of motion:

\sum F=ma

where

\sum F is the net force on the box

m is the mass of the box

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Here we have to consider the horizontal direction, which is the one in which the box is moving. The net force is given by:

\sum F=3.55 N - 0.52 N=3.03 N

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m = 0.200 kg (mass of the box)

Therefore, the acceleration of the box is:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force (also called weight) is the force with which an object is pulled by the Earth towards the Earth's centre.

It is given by

F_g = mg

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m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the acceleration due to gravity

Therefore, the gravitational force on the box is:

F_g=(0.200)(9.8)=1.96 N

c)

The normal force is the force with which a surface pushes back on an object.

For an object lying on a flat surface, there are two forces acting along the vertical direction:

- The gravitational force, F_g, which pushes the object downward

- The normal force, N, which pushes the object upward

As the object is at rest, the vertical acceleration of the object is zero, therefore according to Newton's second law of motion, the net force must be zero:

\sum F=F_g-N=0

Which means that the normal force is equal to the gravitational force:

N=F_g

And so, for the box in this problem, the normal force is

N=1.96 N

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Explanation:

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Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
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Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

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Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

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Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

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