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You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

Another term for technology is

Katyanochek1 [597]

Answer:

a. applied science

Explanation:

$.$

3 0

3 years ago

Read 2 more answers

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

What’s is the relationship between energy and motion ?

LekaFEV [45]

Motion energy is the sum of potential and kinetic energy in an object that is used to do work.

5 0

3 years ago

In this model, the velocity of the spacecraft at position 2 is the velocity of the craft at position 4. at position 1, the direc

Free_Kalibri [48]

1. The velocity of the spacecraft at position 2 is greater than the velocity of the craft at position 4.

This is due the gravity field of the Earth is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft to enter a circular orbit.

2. At point 1, the direction of the spacecraft changes because of the gravitational force between earth and the spacecraft.

As explained in the first answer, this is the exact point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

3. Position 3 represents the orbital path of Earth

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished. If the orbital path of the Earth were the opposite, the effect on the craft would be braking.

Note all of these is related to the gravitational assistance, this consists in a maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe or craft, changing its trajectory.

To learn more about velocity of the spacecraft : brainly.com/question/11900446

#SPJ4

7 0

1 year ago