All categories

4 years ago

A ball was dropped and had a mass of .2 kg and was falling with a force of 2 N, what was its acceleration?

Physics

1 answer:

Musya8 [376]4 years ago

7 0

Answer:

10m/s I believe

Explanation:

F=MA

A=F/M

A=2/0.2

A=10m/s

You might be interested in

What is the international standard for measurement it is a modified version of the metric system

maxonik [38]

Answer:

SI system i think it is right

6 0

3 years ago

What is the relationship between balancing equations and the law of conservation of matter

harkovskaia [24]

Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

7 0

3 years ago

How does mass affect the roller coasters speed?

WARRIOR [948]

More mass, more inertia, less speed, more momentum because momentum is depends its mass and speed. Hope it helps

6 0

3 years ago

Is this right? plzz anwser soon

-Dominant- [34]

Answer:

yes

Explanation:

6 0

3 years ago

Read 2 more answers

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago