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3 years ago

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What is the gravitational potential energy of a rock with the mass of 67 kg if it is sitting on top of a hill .35 kilometers hig

h?

1 answer:

solong [7]3 years ago

8 0

Not to sure but maybe 23.45

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MathPhys Help pls Tysm

HACTEHA [7]

Answer:

8.75

Explanation:

First, find the force of friction.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

3 0

3 years ago

Read 2 more answers

Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0

valentina_108 [34]

Answer:

willie tower

Explanation:

mataas kasi

8 0

2 years ago

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

$p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =$

where:

$M =222u$ is the mass of the original nucleus

$v=420 m/s$ is the initial velocity of the nucleus

$m_1 = 4 u$ is the mass of the alpha particle

$v_1$ is the final velocity of the alpha particle

$m_2 = 222u-4u = 218 u$ is the mass of the daughter nucleus

$v_2 = 350 m/s$ is the final velocity of the nucleus

Solving for $v_1$ , we find the final velocity of the alpha particle:

$v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

4 0

3 years ago

Match the situation with the energy transformation ITEMBANK: Move to Top A boy shooting a rubber band across the classroom A chi

Sonbull [250]

A boy shooting a rubber band across the classroom -->
Elastic potential energy transformed into kinetic energy
<span>The initial energy is the energy stored in the muscles of the boy's arm, which is elastic potential energy. This is converted into motion of the rubber, therefore kinetic energy

A child going down a slide on a playground --> </span>Gravitational potential energy transformed into kinetic energy
On top of the slide, all the energy of the child is gravitational potential energy due to its height with respect to the ground (E=mgh). when it moves down the slide, this is converted into kinetic energy, because the child acquires a speed v (E=1/2 mv^2)
<span>
Rubbing your hands together to warm them on a cold day --> </span>Kinetic energy being transformed into thermal energy <span>
When rubbing hands, we are moving them (kinetic energy), and this energy raises the temperature of the hand's surface (thermal energy)

Turning on a battery operated light --> </span>
Chemical potential energy transformed into radiant energy <span>
A battery works by mean of chemical reactions (chemical potential energy), producing light (so, emitting energy by radiation, i.e. radiant energy)

Using a dc electric motor --> </span> Electrical energy transformed into kinetic energy<span>
A dc electric motor works using currents (so, electrical energy), and the energy produced can be used for example to accelerate a car (kinetic energy)

Using a gas power heater to warm a room --> </span>Chemical potential energy transformed into thermal energy
<span>A gas power heater burns gases (so, chemical reaction, i.e. chemical potential energy) to raise the temperature of the room (thermal energy)

Using a hand crank generator to produce electric current --> Kinetic energy transformed into electrical energy
In a hand-crank generator, the handle is being rotated (kinetic energy) in order to produce an electric current (electrical energy)

Using the light in your room that is plugged into the wall --> </span>Electrical energy transformed into radiant energy
<span>The lamp works by using electrical current flowing into a resistor (electrical energy) and it produces light, so it emits energy by electromagnetic radiation (radiant energy)

</span> <span>

</span>

3 0

3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago