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3 years ago

Which of the following types of energy originates below Earth's surface?

1 answer:

6 0

Geothermal energy. Electromagnetic energy originates at the sun (think UV rays; they are part of the electromagnetic spectrum). Electrical energy technically can be anywhere, as long as there are positive and negative charges. Mechanical energy is like a generator or an engine; there is always something working to get things done/made. I hope this answer and these explanations helped!

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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

What are convection currents and what causes them?

Novay_Z [31]

Caused by the very hot material at the deepest part of the mantle rising, then cooling, sinking again and then heating, rising and repeating the cycle over and over.
It is the movements within the Earth's mantle caused by the heat of the core. The region where two or more tectonic plates meet.

4 0

3 years ago

An unknown mass of each substance, initially at 25.0 ∘C, absorbs 1920 J of heat. The final temperature is recorded. Find the mas

zhannawk [14.2K]

Answer: mass for Pyrex glass 84.21g

mass for sand 61.6g

mass for ethanol 41.32g

mass for water 62.07g

Explanation

By definition specific heat is the amount of heat required to change the temperature of 1 kg mas by 1°C

Q=mcΔT is formula for specific heat

Q is heat transfer

m is mass

ΔT is change in temperature

c is specific heat

c of Pyrex glass= 0.75 j/g°C

c of sand = 0.84 j/g°C

c of ethanol= 2.42 j/g°C

c of water = 4.18 j/g°C

now we will make M(mass) the subject, so equation becomes

m=Q/cΔT

for

pyrex glass Tf=55.4°C

m=1920/(55.4-25)*0.75

m=84.21g {after cutting J(joules) and °C we are left with g(grams)}

for

sand Tf=62.1°C

m=1920/(62.1-25)*0.84

m=61.6g {after cutting J(joules) and °C we are left with g(grams)}

for

ethanol Tf=44.2°C

m=1920/(44.2-25)*2.42

m=41.32g {after cutting J(joules) and °C we are left with g(grams)}

for

water Tf=32.4°

m=1920/(32.4-25)*4.18

m=62.07g {after cutting J(joules) and °C we are left with g(grams)}

i hope you understand the solution, thank you.

7 0

3 years ago

What happens when an object is moved against gravity, such as rolling a toy car up a ramp?

BlackZzzverrR [31]

Answer:

it goes up until we help it to but the moment we stop support it gets affected by gravity and goes back

Explanation:

3 0

3 years ago

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago