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4 years ago

Please need help on this

Physics

1 answer:

Rasek [7]4 years ago

7 0

Im pretty sure this is A. i may be incorrect, give brainliest if im correct please!

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When can hfc-134a become combustible in a mobile air conditioning system?

zubka84 [21]

The mixture of HFc-134a with high concentration of air at raised pressure can become combustible in presence of a ignition source, that is oxygen enriched environment.Thus at high pressure in a mobile air condition led to combustible of HFC-134a.

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4 years ago

An object with a mass of 20kg is pushed with a force of 60 Newtons(N). What is the object’s acceleration?'

jenyasd209 [6]

Answer: $a=3m/s^2$

Explanation:

$Formula: F=m*a$

Where;

F = force

m = mass

a = acceleration

Solve for a;

$a=\frac{F}{m}$

$a=\frac{60N}{20kg}$

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4 years ago

Read 2 more answers

Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o

likoan [24]

Answer:

1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

$E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]$

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

$E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\$

And the kinetic energy will be:

$E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]$

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

$E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]$