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Free_Kalibri [48]

3 years ago

9

if the Earth's core provides the heat that drives plate tectonics, then what will eventually happen as the Earth's core cools do

wn over billions of years?

1 answer:

Eduardwww [97]3 years ago

6 0

If the Earth's core cools down, then one of the first things that would happen is that plate tectonic activity would slow down. Over time, the Earth would then dry up into this ball of dry soil because the heat of the Earth's core drives water up to the surface. Since the crust would be void of majority of liquid, it would shrink by an unknown amount.

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Can someone please help me with this one

Brilliant_brown [7]

Answer:

in left

Explanation:

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5 0

3 years ago

The force of gravity is an inverse square law. this means that, if you double the distance between two large masses, the gravita

Nadya [2.5K]

F~1/r²
weakens by 1/4

4 0

3 years ago

The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o

sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, $P_f=20\ J$

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

$P_i=K_f+P_f$

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20$

$\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20$

$50-20=\dfrac{1}{2}mv^2$

$30=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{60}{100}}$

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0

3 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

3 years ago

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

4 0

3 years ago