All categories

2 years ago

The cycle of day and night is a result of Earth’s spinning on its axis, which is Earth’s .

Physics

1 answer:

nekit [7.7K]2 years ago

5 0

Answer:

Rotation

Explanation:

i did the test Gl.

You might be interested in

What would you use to to measure the weight of a truck?? I’m needing a word that is 8 letters long

myrzilka [38]

Answer:

Railroad Scale?

Explanation:

5 0

2 years ago

A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?

kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

= 1092 ft³

The weight of the air = density x volume

= 0.07 lbs/ft³ x 1092 ft³

= 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0

2 years ago

A medieval instrument used to determine the position of the sun:

Stels [109]

Answer: astrolabe

Explanation:

4 0

2 years ago

Which of the following is a heterogeneous mixture .

lianna [129]

Answer:

chicken noodle soup

Explanation:

Heterogenous mixture is the mixture where you can see differnt parts of it, hence "hetero".

you can see chicken, noodles, and broth in chicken noodle soup

5 0

2 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

3 years ago