OOF it is the same for both

Name the forces acting on a plastic bucket container water held above ground level in your hand. Discuss why the forces acting o

SashulF [63]

There are two forces at play:

The gravitational force acting downward due to the mass of the bucket and the water that it contains.
The upward force that your hand exerts on the bucket.

If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.

Having 0 net force means the bucket doesn't undergo any acceleration, or change in motion.

5 0

3 years ago

An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the

Bingel [31]

Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision

Explanation:

The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:

$\Delta p = F \Delta t$

where

$\Delta p$ is the change in momentum

F is the average force

$\Delta t$ is the duration of the collision

In this problem, neither side of the equation can be used to measure the change in momentum. In fact:

- The change in momentum (left side) is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

u is the initial velocity

v is the final velocity

Here the final velocity is not known, so it's not possible to use this side of the equation

- The impulse (right side) is given by

$F\Delta t$

here the average force is known, however the duration of the collision is not known, so it's not possible to use this side of the equation.

Learn more about momentum:

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago

A particle travels in a circle of radius 76 cm and completes one revolution in 4.5 s. What is the centripetal acceleration of th

Dimas [21]

r = radius of the circle traveled by the particle = 76 cm = 0.76 m

T = time period of revolution for the particle = 4.5 s

w = angular velocity of the particle

angular velocity of the particle is given as

w = 2π/T

inserting the values

w = 2 (3.14)/4.5

w = 1.4 rad/s

a = centripetal acceleration of the particle in the circle

centripetal acceleration is given as

a = r w²

inserting the values

a = (0.76) (1.4)²

a = 1.5 m/s²

3 0

3 years ago

Read 2 more answers

What are the disadvantages and advantages for electromagnetic spectrum

Basile [38]

Answer:

Advantages: In small quantities they help your body produce vitamin D.

Disadvantages: They can cause sunburn or even skin cancer if used in large quantities.

Explanation:

3 0

3 years ago

A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0

2 years ago