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3 years ago

PE=30J, m=?, g=10m/s2, h=10m

1 answer:

5 0

Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

$PEgrav=mgh$
$30J=m(10)(10[tex] \frac{30}{100} =m$ )[/tex]

Do operations that you can with what is given first.

$30J=m(100m)$

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

$\frac{30}{100} =m$

m = 0.30kg

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The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa

xxMikexx [17]

Answer:

4.16 L

Explanation:

Assuming constant temperature,

At the edge of Typhoon Odessa: P₁ = 1 atm = 1013.3 mbar,

V₁ = 4.0 L

At the center of Typhoon Odessa: P₂ = (1013.3 - 40.0) mbar = 973.3 mbar

V₂ = ? L

For a fixed amount of gas at constant temperature (Boyle's law) : P₁V₁ = P₂V₂

V₂ = V₁ × (P₁/P₂)

V₂= (4.0) × (1013.3/973.3)

V₂= 4.16 L

7 0

3 years ago

Peter notices that his eight-year-old daughter frequently mentions the unusual habits of a new boy, who is from a different cult

pickupchik [31]

The answer should be C. Encourage his daughter to make friends with the new boy.

Hope this helps and have a happy new year

6 0

3 years ago

URGENT!!!!!!!<br><br> PLEASE HELP WITH THIS PHYSICS PROBLEM

Levart [38]

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$

3 0

3 years ago

If a car's speed triples, how does the momentum and kinetic energy of the

Harrizon [31]

Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.

Explanation:

The momentum of a car (an object) is

p= mv

where

m is =the mass of the object( in this case car)

v is its= velocity

While the kinetic energy is is given by the formulae

K=1/2mv²

To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,

v¹= 3v

So that the momentum change becomes

p¹=mv¹=m (3v)= 3mv

mv=p

therefore p¹= 3p

we can see that the momentum also triples.

And the kinetic energy change becomes

K¹=1/2m(v¹)²= 1/2m (3v)²

= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²

1/2mv²=K

K¹= Kinetic energy = 9k

but Kinetic energy increases 9 times

7 0

3 years ago

When there's a hazard ahead, it's almost always quicker for you to _________ than to come to a full stop.

TEA [102]

When there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

<h3>What is an hazard?</h3>

Hazard refers to any obstacle or other feature which causes risk or danger.

Living organisms respond to hazards via the production of adrenaline hormone. This hormone causes a flight response away from the hazard.

Therefore, when there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

Learn more about hazards at: brainly.com/question/5338299

5 0

2 years ago