Which two statements about an electric motor are true

an 80 kg cart goes around the inside of a vertical loop of a roller coaster. the radius of the loop is 5m and the cart moves at

GarryVolchara [31]

Answer:

F= 224 N

Explanation:

Given that

mass ,m = 80 kg

Radius ,r= 5 m

speed at the top v= 8 m/s

The force at the top = F

Now by using the Second law of Newton's

$F+mg=\dfrac{mv^2}{r}$

Now by putting the values

$F+80\times 10 = \dfrac{80\times 8^}{5}$

Take g = 10 m/s²

$F= \dfrac{80\times 8^2}{5}-800$

F= 224 N

Therefore the force exerted by the track at the top position will be 224 N.

3 0

4 years ago

During a supernova, the outer layers of a star are blown off and the star's core shrinks down by a factor of 10,000 or more to f

Lana71 [14]

Answer:

a. the core will spin faster.

Explanation:

By law of conservation of angular momentum

(mvR)i= (mvR)f

m= mass of star

v= speed of star

R= radius of star

i= initial

f= final

since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.

Hence, a. the core will spin faster.

8 0

4 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s