All categories

3 years ago

PLS HELP ASAP PLEASE PLEASE PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Physics

1 answer:

7nadin3 [17]3 years ago

7 0

Im pretty sure the answer is d

You might be interested in

Another name for a period is a family?

Zepler [3.9K]

If you are referring to science, then you are correct. A period is also known as a family.

8 0

3 years ago

Examples of applied force

Vladimir [108]

Answer:

Push - The most common form of force is a push through physical contact (like a lawnmower or shopping cart)

Pull - You can apply a force by directly pulling on an object (like pulling a wagon)

Explanation:

4 0

3 years ago

Read 2 more answers

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

4 years ago

suppose you needed to build a raft to cross a fast-moving river. Describe the physical and chemical properties of the raft that

erastovalidia [21]

1.Density of the material building the raft must be lower than the water.
2. Material must not react with water. 
3.Material must have high strength. 
4.Raft must be wide in-order to avoid drawing in the river.

4 0

3 years ago

The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi

SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

$\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3$

$\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3$

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0

3 years ago