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fgiga [73]
3 years ago
7

A ball bounced by a basketball player on the floor bounces back up at her. Newton's First Law Newton's Second Law Newton's Third

Law
Physics
1 answer:
taurus [48]3 years ago
4 0

Answer:

Newton's Third Law of Motion

Explanation:

Newton's Third Law of Motion which states that, for every action there is an equal but opposite reaction.

This ultimately implies that, in every interaction, there is a pair of forces acting on the two interacting objects.

In this scenario, a ball bounced by a basketball player on the floor bounces back up at her.

According to Newton's Third Law of Motion, the statement above simply means that in every interaction, there is a pair of forces acting on the two interacting objects i.e the ball and floor. The size of the force on the ball equals the size of the force on the floor. These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

Hence, the ball bounced by the basketball player on the floor would bounce back in equal magnitude.

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Why are not all microorganism harmful?
son4ous [18]


In nature there are two categories of microorganisms as relating to health. Microorganisms that are considered harmful to humans are called pathogens and these cause disease. Examples include bacteria such as streptococcus which cause sore throat and salmonella which cause typhoid disease.

 There are some microorganisms which are helpful to man and they live mostly on the skin of man or in his gut and are mostly bacteria. They are collectively called bacterial normal flora.

In man the normal bacterial flora of the skin include staphylococcus found on dry skin, cornybacteria found in moist skin sites and propionibacteria in the sebaceous sites (head, neck, trunk) of the body. Normal bacterial flora of the gut include Escherichia coli.

One of the major function of bacterial flora is actually to protect our bodies  by competing for space with pathogens preventing them from gaining a foothold in our bodies.



8 0
3 years ago
At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either
GrogVix [38]

Answer:

The direction of the force at A and B is perpendicular to the walls of the container.

The direction of the force at C is down.

The direction of the force in D is up

The direction of the force at E is to the left.

The attached figure shows the forces exerted by the water at points A, B, C, D and E.

Explanation:

The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.

The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.

The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.

The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.

4 0
3 years ago
The gravitational force between two objects that
leonid [27]

Answer:

The answer to your question is    m₂ = 38.5 kg

Explanation:

Data

distance = d = 2.1 x 10⁻¹ m

Force = 3.2 x 10⁻⁶ N

m₁ = 55 kg

m₂ = ?

G = 6.67 x 10 ⁻¹¹ Nm²/kg²

Process

1.- To solve this problem use Newton's law of Universal Gravitation.

             F = G m₁m₂ / r²

-Solve for m₂

            m₂ = Fr² / Gm₁

2.- Substitution

            m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)

3.- Simplification

            m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹

4.- Result

            m₂ = 38.5 kg

5 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
5 0
2 years ago
g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron
Butoxors [25]

Answer:

The value is v  =  2.999 *10^{8} \  m/s

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is t = 189 \ light-years

    The  time to be spent on the ship is  t_{s} =  12 \  years

Generally speed can be obtained using the mathematical relation represented below

       t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }

=>     v  =  2.999 *10^{8} \  m/s

5 0
3 years ago
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