what type of energy is kinetic energy? A nuclear energy B chemical energy C potential energy D mechanical energy

A cat weighing 7 kg chases a mouse at a speed of 4 m/s. What is the kinetic energy of the cat?

jeka57 [31]

54J
You calculate kinetic energy by the equation.
KE= 1/2mv^2
m= mass
v= velocity
then we have
KE = 1/2 x 3 x 6^2
KE = 1/2 x 3 x 36 = 54J

8 0

3 years ago

An empty truck and a fully loaded truck are next to each other at the traffic light. When the light turns green the lighter truc

Lilit [14]

This is correct. The lighter truck moves faster.

4 0

3 years ago

Please help meeeee will mark as brainliest

Katyanochek1 [597]

Answer:

a) pressure means force acted on a body per unit area

6 0

3 years ago

Which image represents the force on a positively charged particle caused by an approaching magnet?

Amanda [17]

Answer:

Image B represents the force on a positively charged particle caused by an approaching magnet.

Explanation:

The most fundamental law of magnetism is that like shafts repulse each other and dissimilar to posts pull in one another; this can without much of a stretch be seen by endeavoring to put like posts of two magnets together. Further attractive impacts additionally exist. On the off chance that a bar magnet is cut into two pieces, the pieces become singular magnets with inverse shafts. Also, pounding, warming or winding of the magnets can demagnetize them, on the grounds that such dealing with separates the direct game plan of the particles. A last law of magnetism alludes to maintenance; a long bar magnet will hold its magnetism longer than a short bar magnet. The domain theory of magnetism expresses that every single enormous magnet involve littler attractive districts, or domains. The attractive character of domains originates from the nearness of significantly littler units, called dipoles. Iotas are masterminded in such a manner in many materials that the attractive direction of one electron counteracts the direction of another; in any case, ferromagnetic substances, for example, iron are unique. The nuclear cosmetics of these substances is with the end goal that littler gatherings of particles unite as one into zones called domains; in these, all the electrons have the equivalent attractive direction.

4 0

3 years ago

2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the

frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum = 2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = $\frac{1}{2}$ mv²

v² = 2gh₁

v = √(2gh₁)

from the second equation; s = ut + $\frac{1}{2}$ at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{3}$ mR²) ω²

v = √( $\frac{6}{5}$ gh₁ )

so we substitute √( $\frac{6}{5}$ gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$ gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0

3 years ago