acceleration is defined as the rate of change for which characteristic? a. displacement b. position c. velocity d. time

Which equation represents an equilibrium system??

MrRissso [65]

A. 2 S O 2, gas, plus O 2, gas, in equilibrium with 2 S O 3, gas.

7 0

4 years ago

When we use an analogy that represents the expanding universe with the surface of an expanding balloon, what does the inside of

svetlana [45]

Answer: When we use an analogy that represents the expanding universe with the surface of an expanding balloon, what does the inside of the balloon represent? The inside of the balloon does not represent any part of our universe.

7 0

3 years ago

A car is traveling at a speed of 54 km/h. Breaks are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how far

Lena [83]

Explanation:

u=54 km/h

54*5/18=15 m/s

v=0m/s

t=?

acceleration=-0.5m/s^2

we know that a=v-u/t

so,

t=v-u/a

t=15-0/0.5

=15/0.5

=30

therefore, the time is 30 second

Hope this answer helps you..

8 0

3 years ago

PLEASE HELP!! I’ll give brainliest pls

marin [14]

Answer:

A

Explanation:

houses use alternating current source

6 0

3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago