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2 years ago

What are the forces that act on the ball pushed on the floor?Plss answer

Physics

1 answer:

vaieri [72.5K]2 years ago

4 0

Explanation:

The ball, for example, will feel gravity pulling it downward and the ground pushing it upward in the direction it is rolling. (Add this if the ball is rolling on the floor.) Friction is the force that causes the ball to slow down because it acts in the opposite direction that it is moving.

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What would be the y-component of a velocity vector that had a

valentina_108 [34]

Answer:

18.5 m/s

Explanation:

Just did it on EdPuzzle

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3 years ago

The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}

tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

$K.E=\frac{1}{2}mv^2$

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2$

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2$

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

$\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}$

Given that, v = 2v'

$\frac{K.E}{K.E'}=\frac{1}{1}$

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0

3 years ago

How can density be determined in a lab of a rectangular solid?

True [87]

We will first record its mass and then its volume by measuring its dimensions
then divide mass by volume and will get density of regular solid

6 0

3 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

3 years ago

Read 2 more answers

You carry a 5 kg sack of potatoes across the store for 5 minutes, and you wait in line holding it for 30 seconds before you get

ludmilkaskok [199]

Answer:

The only work done is when the person lifts the sack over a distance, W = 78.48 [N]

Explanation:

We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.

W = F * d

where:

F = force [N]

d = distance [m]

The force is given by the producto of the mass by the gravity.

F = 5 * 9.81 = 49.05 [N]

W = 49.05 * 1.6 = 78.48 [N]

3 0

4 years ago

What are the forces that act on the ball pushed on the floor​?Plss answer

What are the forces that act on the ball pushed on the floor?Plss answer