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3 years ago

The form of energy in fossil fuels is what kind of energy?

Physics

2 answers:

Nonamiya [84]3 years ago

7 0

Solar energy or sunlight...

kupik [55]3 years ago

6 0

It's thermal energy.

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What does this image reveal about gravityon the moon compared to Earth?

elena-14-01-66 [18.8K]

ANSWER:

The Moon's gravity is less than Earth's.

STEP-BY-STEP EXPLANATION:

When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.

The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.

As the gravity is less, you can do things such as the one shown in the image.

As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.

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1 year ago

Que es entrenamientos físicos?

Gnesinka [82]

el uso sistemático de ejercicios para promover la aptitud física y la fuerza.

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3 years ago

It takes a slug 20 minutes to travel from the grass to the trash can , a trip of 15 meters. How far could the slug travel in 60

inn [45]

Answer:

45 meters

Explanation:

20 min = 15 meters

So if 20 x 3 = 60

you have to do 3 x 15 !

- which equals to 45 <3

<u>- mark me brainlest pls . </u>

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4 years ago

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$

(d) The acceleration vector stays constant throughout, so

$\mathbf a(t)=\boxed{-g\,\mathbf j}$

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3 years ago

HELP PLEASE ASAP!!!<br> I CAN'T FAIL PHYSICS<br> xoxo thank you!

Allushta [10]

They are malleable and lustrous, and can conduct both electricity and thermal heat

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3 years ago