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lora16 [44]
3 years ago
6

What happens to a football thrown in space ,and why

Physics
2 answers:
Neporo4naja [7]3 years ago
8 0

hdfjfAnswer:

Explanation:

nbdhfb

Dafna11 [192]3 years ago
7 0

Answer:

<h2>In deep space, far enough away from large objects like stars and planets that gravity may be neglected, a thrown ball will travel in a straight line and the person throwing the ball will also travel in a straight line in the opposite direction.</h2>

Explanation:

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Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig
MissTica

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

         

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

6 0
3 years ago
How to derive the fourth equation of motion?
Leya [2.2K]

Answer:

To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

Explanation:

v= u + at

s =( u + v ) t /2

s = ut + at2/2

v2 = u2 + 2as

6 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

6 0
3 years ago
Is the frictional force the same as the applied force when the net force equals zero?
DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

6 0
3 years ago
A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux
bixtya [17]

The total flux through the cylinder is zero.


In fact, the electric flux through a surface (for a uniform electric field) is given by:

\Phi = E A \cos \theta

where

E is the intensity of the electric field

A is the surface

\theta is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.


We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because \theta=90^{\circ} and \cos \theta=0).


On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

\Phi_1 = E \pi r^2

where r is the radius, and where we have taken \theta=0^{\circ} since the perpendicular to the surface is parallel to the direction of the electric field, so \cos \theta=1. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so \theta=180^{\circ} and \cos \theta=-1, therefore the flux is

\Phi_2 = -E \pi r^2

And the net flux through the cylinder is

\Phi = \Phi_1 + \Phi_2 = E \pi r^2 - E \pi r^2=0

4 0
3 years ago
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