All categories

2 years ago

You dribble a basketball while walking on a basketball court list and describe 3 pairs of action-reaction forces

Physics

1 answer:

elena-s [515]2 years ago

5 0

Answer:

1. Walking

2. Pushing the basketball with hand

3. Ball bouncing off the ground

Explanation:

As per Newton's third law of motion every action has an equal and opposite reaction. Looking at the given scenario following are the pairs of action-reaction forces:

1. Walking: When you walk, you apply action and push the ground and the ground pushes you back with the same force.

2. Pushing the basketball with hand: when you push the ball that is action, the ball also applies force on your hand which is the reaction.

3. Ball bouncing off the ground: When ball hits the ground, it applies force i.e. action and the ground pushes it back with the same force which is the reaction.

You might be interested in

While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the

Vladimir [108]

24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²

4 0

2 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

2 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>

5 0

2 years ago

which type of front occurs when cold air and warm air are next to each other but are at a a stand still ?

algol13

A stationary front
hope this helps

3 0

2 years ago

Read 2 more answers