Which of the following is a theory stating that one plate is forced beneath another plate?

What happens when you "crack/pop your knuckles?

Art [367]

It could result in it not being good for your joints, as well as in the long run but shouldn't cause problems when your a child. I hope this helps your question!

6 0

3 years ago

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the

mash [69]

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

4 0

4 years ago

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

tester [92]

Answer:

a) I = 1,75 10-² kg m² and b) I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

I = I core + I shell

The moment of inertia of a solid sphere is

I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

R = d / 2

R= 0.196 m / 2 = 0.098 m

I core = 2/5 1.6 0.098²

I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

R = 0.206 / 2

R = 0.103 m

I shell = 2/3 1.6 0.103²

I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

I = I core + I shell

I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

I = 17.47 10⁻³ kg m²

I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

R = 0.216 / 2

R = 0.108 m

It is a uniform sphere

I = 2/5 M R²

I = 2/5 3.2 0.108²

I = 1.49 10⁻² kg m²

7 0

3 years ago

The gravitational field at the Moon in N/kg due to the Earth is approximately (G = 6.67 × 10-11 N m2/kg2, the mass of the Earth

rosijanka [135]

Answer:

F = 2.69 10⁻³ m [ N]

Explanation:

This exercise asks to calculate the gravitational field of the Earth on the lunar surface, let's use the universal gravitation law

F = G m M / r²

where m is the mass of the body, M the mass of the Earth and r the distance between the Earth and the Moon

F = (G M / r²) m

F = (6.67 10⁻¹¹ 5.98 10²⁴ / (3.85 10⁸)² ) m

F = 2.69 10⁻³ m [ N]

This force is directed from the Moon towards the Earth, therefore it reduces the weight of the body

8 0

3 years ago

A student used apparatus as shown above. The beaker held 750 g of a liquid. The current from the power supply was 1.8 A and the

liq [111]

Answer:

$2365.71\ \text{J/kg}^{\circ}\text{C}$

Explanation:

V = Voltage = 230 V

I = Current = 1.8 A

$\Delta T$ = Temperature change = $(40-12)^{\circ}\text{C}$

t = Time taken = 2 minutes

m = Mass of liquid = 750 g

c = Specific heat capacity of the liquid

Energy required to heat the water is equal to the heat released due to current passing

$mc\Delta T=VIt\\\Rightarrow c=\dfrac{VIt}{m\Delta T}\\\Rightarrow c=\dfrac{230\times 1.8\times 2\times 60}{0.75\times (40-12)}\\\Rightarrow c=2365.71\ \text{J/kg}^{\circ}\text{C}$

The specific heat capacity of the liquid is $2365.71\ \text{J/kg}^{\circ}\text{C}$

6 0

3 years ago