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QveST [7]
3 years ago
6

Question 5 of 10 Which two factors affect the amount of thermal energy an object has? A. The amount of space between its particl

es B. The mass of the object C. The amount of motion its particles have D. The directions in which its particles are moving​

Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer: A: The Mass of Object AND D. The amount of motion it’s particles have

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Physics Homework
katrin2010 [14]

Explanation:

a. Average speed = distance / time

= 100 m / 70 s

= 1.43 m/s

b. Average displacement = displacement / time

= 0 m / 70 s

= 0 m/s

Distance is the length of the path traveled.  Displacement is the difference between the final position and initial position.

4 0
4 years ago
Upwelling _____.
Assoli18 [71]
- - A up welling is a solution appearing and moving towards the ocean surface meaning as, its going up,
Therefore, C <span>happens when water rises to replace moving surface water</span>
8 0
4 years ago
Read 2 more answers
An airplane flies at 400 km/h in a 100km/h hurricane crosswind. Find the resultant speed of the plane.
kow [346]
I think the airplan flies in same speed 400
5 0
4 years ago
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a
Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

w_1g = 1.4*9.81*y*sin30 = - 6.867y

Workdone by gravity on the mass of the suspended block is:

w_2g = 0.06*9.81 = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y²  - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y²  - 6.867y + 0.5886 = 0.0588

-20 y²  - 6.867y + 0.5886 - 0.0588

-20 y²  - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y²  + 6.867y - 0.5298 = 0

Using the quadratic formula:

\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

where;

a = 20 ; b = 6.867 c= - 0.5298

\dfrac{-(6.867)  \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}

= \dfrac{-(6.867)  + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \  \dfrac{-(6.867)  - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}= 0.0649 OR  −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0
4 years ago
Given your choice for part 1, would it be possible, in principle under ideal circumstances, for a balloon to stick to a wall ind
podryga [215]

Answer:

Explanation:1)Rub a balloon on your head and stick it to the wall. And it sticks—very nice. We’ve

talked in lecture about why it sticks. This question is about why, after a while, it fall off.

Consider each of the following explanations.

i) It falls because it uses up the energy you put on it when you rubbed it.

ii) It falls because it uses up the charge you put on it when you rubbed it.

iii) It falls because the energy you put on it slowly leaks off into the wall or the air.

iv) It falls because the charge you put on it slowly leaks off into the wall or the air.

Answer: It falls because the charge you put on it slowly leaks off into the wall or the air.

2) Given your choice for part a, would it be possible, in principle under ideal

circumstances, for a balloon to stick to a wall indefinitely?

Answer: If you can keep the charge from leaving the balloon, And if the

wall is a very good insulator also this will prevent the ball from falling.

8 0
3 years ago
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