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3 years ago

Plz help

Physics

2 answers:

Vesna [10]3 years ago

8 0

Upper mantle durrrr it’s cuz it causes movement cuz like the heat or something liek that

Gennadij [26K]3 years ago

3 0

Upper mantle

The theory of the plate tectonics uses the convection cells in the upper mantle layer as the foundation for the movement of the continents.
To put it simple the process is explained like the convection cells create a lot of pressure, that pressure, supported by the high temperatures and magma, influence the crust above, and manage to crack it on certain places. As the crust cracks, it becomes a separate entity, and under the enormous force from bellow it slowly moves into a particular direction, and that is actually the movement of the tectonic plates.

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Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plate

Alecsey [184]

Answer:

The magnitude of the new electric field is <u>35820 N/C</u>.

Explanation:

Given:

Original magnitude of electric field (E₀) = 2388 N/C

Original voltage = 'V' (Assume)

Original separation between the plates = 'd' (Assume)

Now, new voltage is three times original voltage. So, $V_n=3V$

New distance is 1/5 the original distance. So, $d_n=\dfrac{d}{5}$

Now, electric field between the parallel plates originally is given as:

$E_0=\frac{V}{d}=2388\ N/C$

Let us find the new electric field based on the above formula.

$E_n=\frac{V_n}{d_n}\\\\E_n=\frac{3V}{\frac{d}{5}}\\\\E_n=15(\frac{V}{d})$

Now, $\frac{V}{d}=2388\ N/C$ . So,

$E_n=15\times 2388=35820\ N/C$

Therefore, the magnitude of the new electric field is 35820 N/C.

3 0

3 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

Which observation is the best evidence that some colors of visible light are

Effectus [21]

Answer:

id say D

Explanation:

because its the only answer that has to do with colors and visable light

but A and B also have to do with eachother so i feel like it could be with one of those but it def isnt C

I am sorry if i am wrong, let me know tho!

8 0

3 years ago

Read 2 more answers

Can you please answer theese two qeastion

wel

The first one is C!

I hope the answer helped! please mark brainliest and vote 5 stars! >:)

6 0

3 years ago

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Paha777 [63]

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

$P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s$

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

$P_1=P_2$

$0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2$

$v_2=1 m/s$

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

4 0

4 years ago