A farmer heaves a 7.56 kg bale of hay with a final velocity of 4.75. What is the kinetic energy of the bale?

A man ties one end of a strong rope 8.17 m long to the bumper of his truck, 0.524 m from the ground, and the other end to a vert

Kamila [148]

Answer:

2442.5 Nm

Explanation:

Tension, T = 8.57 x 10^2 N

length of rope, l = 8.17 m

y = 0.524 m

h = 2.99 m

According to diagram

Sin θ = (2.99 - 0.524) / 8.17

Sin θ = 0.3018

θ = 17.6°

So, torque about the base of the tree is

Torque = T x Cos θ x 2.99

Torque = 8.57 x 100 x Cos 17.6° x 2.99

Torque = 2442.5 Nm

thus, the torque is 2442.5 Nm.

8 0

3 years ago

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = $\frac{3}{1000}$ kg

Therefore F = 0.003×5 = 0.015 m/s2

3 0

3 years ago

The sum of kinetic energy of molecules in a body is ______

Margarita [4]

Mechanical energy is the answer

6 0

2 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago

Refer to science 10 (mirror mirror on the wall)how do the height and width of the object compare with the height and width of th

Irina-Kira [14]

Depends on what type of mirror that is. I am going to assume this is a plain mirror (from the phrase), which means the height and width of the object and image is exactly the same.

6 0

3 years ago