2 years ago

Which of the following are appropriate acceleration units?

Physics

1 answer:

Paraphin [41]2 years ago

8 0

Answer:

meters per second squared is the appropriate units for acceleration units.

You might be interested in

All of the fallowing are possible sources of error in a scientific investigation exept for

Whitepunk [10]

c adding research resources during an investigation

6 0

3 years ago

A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c

Y_Kistochka [10]

Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water

Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)

Answer: 44,100 J

3 0

2 years ago

Which of these actions will increase friction? Select three options. Scratching a surface to make it rougher polishing a surface

k0ka [10]

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

5 0

2 years ago

Read 2 more answers

If are spaced closely together on the map,there is a drastic temperature change over the distance

AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0

3 years ago

Read 2 more answers

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago