When a body falls freely under gravity, then the work done by the gravity is ___________

Answers

Answer 1

Answer:

Answer: positive

Explanation:

Gravity can be defined as the force with which the body is attracted towards the center of the earth, or towards any other body. If the force acting on the body is in the direction of displacement then the word done by the applicable force is positive. This causes the free fall of the ball under the influence of gravity is also positive.

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A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
A missile is moving 1350 m/s at a 25.0 angle
Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.2. It is moving to the right with a net force of 10 N.3. It is in dynamic equilibrium with a net force of 0 N.4. It is in static equilibrium with a net force of 0 N.
In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 28.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe

A stream of water emerging from a faucet narrows as fails. The cross-sectional area of the soutis As -6.40 cm. water comes out of the spout at a speed of 33.2 cm/s, and the waterfalls h = 7.05 cm before iting the bottom of sink What is the cross-sectional area of the water stream just before it is the sink? a. 0.162 cm3 b. 1.74 cm3c. 6.21cm3d. 0.943cm3

Answers

Answer:

The area of the water stream will be 1.74 cm^2

Explanation:

initial velocity of water u = 33.2 cm/s

initial area = 6.4 cm^2

height of fall = 7.05 cm

final area before hitting the sink = ?

as the water falls down the height, it accelerates under gravity; causing the speed to increase, and the area to decrease.

first we find the velocity before hitting the sink

using

$v^(2) = u^(2) + 2gh$ -----Newton's equation of motion

where v is the velocity of the water stream at the sink

u is the initial speed of the water at the spout

h is the height of fall

g is acceleration due to gravity, and it is positive downwards.

g = 981 cm/s^2

imputing relevant values, we have

$v^(2) = 33.2^(2) + (2 * 981 * 7.05)$

$v^(2) = 1102.24 + 13832.1 = 14934.34$

$v = √(14934.34)$ = 122.206 cm/s

according to continuity equation,

A1v1 = A2v2

where A1 is the initial area

V1 = initial velocity

A2 = final area

V2 = final velocity

6.4 x 33.2 = 122.206 x A2

212.48 = 122.206 x A2

A2 = 212.48 ÷ 122.206 ≅ 1.74 cm^2

A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?

Answers

Answer:

Time = 1.61 seconds

Explanation:

Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds

A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block

Answers

Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77

What is Friction?

According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.

F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.

Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.

To learn more about Friction, refer to the link:

brainly.com/question/13000653

#SPJ2

Answer:

7.77

Explanation:

F=μ*m

n=w which also means n=mg

14.7=0.193*n

n=76.2

76.2=m*9.8

m=7.77

You observe three carts moving to the right. Cart A moves to the right at nearly constant speed. Cart B moves to the right, gradually speeding up. Cart C moves to the right, gradually slowing down. Which cart or carts, if any, experience a net force to the right

Answers

Answer:

Explanation:

Cart A is moving to the right with constant speed i.e. net acceleration is zero

because acceleration is change in velocity in given time

Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s

This indicates the net force acting on the cart towards right

For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answers

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = $5.0* 10^(- 5) T$

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

$B = \farc{\mu_(o)I}{2\pi d}$

where

d = distance from current carrying wire

Now,

$d = (\mu_(o)I)/(2\pi B)$

$d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))$

d = 0.02 m 2 cm

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Answers

Answer:

F = 10.8N

Explanation:

Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t =0.5s

From Newtown's second law of motion the average force can be found. This law states that the product of the force experienced by a body and the time t of the force acting on the body is equal to the change in momentum of the body. Mathematically it can be stated as follows

F×t = m(v2 – v1)

F = m(v2 – v1)/t = 0.4(25 – 12)/0.5 = 10.8N

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