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Firlakuza [10]
3 years ago
7

A ball is thrown from the ground with nonzero horizontal and vertical initial velocities. While the ball is in the air, assuming

that only the force of gravity acts on it, which of the following statements is true? a. Both the horizontal and vertical components of momentum are constant. b. Only the horizontal component of momentum is constant. c. Neither the horizontal nor the vertical components of momentum are constant.
Physics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

B

Explanation:

Solution:-

- Take a coordinate system as follows:

                   x: Directed along the ground

                   y: Directed vertical to ground

- We will assume that the initial vertical and horizontal non-zeroes velocities are given as follows:

                          v_x_i = v_o_x\\\\v_y_i = v_o_y\\

- After a ball is thrown it continues a path of parabolic projectile. The motion of the ball can be analyzed in each coordinate system. We will assume that effects of air-resistance are negligible.

- Therefore, only gravity acts on the ball in the vertical direction. We can use kinematic equation of motions to determine the velocity of ball in either ( x-y ) direction at any instant of time ( t ).

- Use first kinematic equation of motion in both x and y directions.

                         v_x_f = v_o_x + a_x*t\\\\v_y_f = v_o_y + a_y*t\\

- The accelerations ( ax and ay ) in the direction of each axis are to be determined. We know that the gravity acceleration ( g ) acts in vertical direction or along y-axis ( ay ) and always directed downwards while velocity is directed up. Since, we neglected the effects of air-resistance there is no acceleration in the x-direction ( ax = 0 ) .

                        v_x_f = v_o_x\\\\v_y_f = v_o_y - gt\\

- We see that the horizontal velocity of the ball ( vxf ) at any point in time remains equal to the initial horizontal velocity; hence, it is constant throughout the journey.

- However, the velocity of the ball in vertical direction( vyf ) is changing for every unit of time ( t ) under the influence of gravitational acceleration. Hence, it is not constant throughout the journey

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I need help understanding this concept . Would appreciate it so much. Thank you
Firlakuza [10]

Answer:

Explanation:

Both these questions are based on the Universal Law of Gravitation, which is given by :

F = Gm1m2 / r²

2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5

   F = 6.67 x 10⁻⁵ x 8 / 1.5

   F = 35.57 x 10⁻⁵ N

3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴

   F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29

   F = 63.13 x 10⁻³ N

7 0
2 years ago
How did you manage to overcome the difficulties that arose?
Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0
2 years ago
A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds
horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

4 0
3 years ago
Read 2 more answers
Which process produces the energy radiated by the star when it becomes a main sequence star?
inysia [295]
The process that produces the energy radiated by stars is nuclear fusion in the core.
For a star on the main sequence, it's the fusion of hydrogen nuclei into helium.
8 0
3 years ago
A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr
astraxan [27]

Answer:

x =  0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2

Now, we will calculate the force on the piston due to atmospheric pressure:

Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N

Now, for the compression of the spring we will use Hooke's Law as follows:

F = kx\\

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

<u>x =  0.0734 m = 7.34 cm</u>

8 0
3 years ago
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