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svlad2 [7]
3 years ago
15

At which of the above boundries is sea floor destroyed

Physics
1 answer:
Slav-nsk [51]3 years ago
7 0
The continental crust
Hope it helps!!
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A cube of water 10 cm on a side is placed in a microwave beam having Ea = 11 kV/m‘ The microwaves illuminate one faceofthe cube,
Shtirlitz [24]

Answer:

The time is 133.5 sec.

Explanation:

Given that,

One side of cube = 10 cm

Intensity of electric field = 11 kV/m

Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.

We need to calculate the rate of energy transfer from the beam to the cube

Using formula of rate of energy

P=(0.80)IA

P=0.80\times\dfrac{c\mu_{0}E^2}{2}\times A

Put the value into the formula

P=0.80\times\dfrac{3\times10^{8}\times8.85\times10^{-12}\times(1.1\times10^{4})^2}{2}\times(10\times10^{-2})^2

P=1285.02\ W

We need to calculate the amount of heat

Using formula of heat

E =mc\Delta T

E =\rho Vc\Delta T

Put the value into the formula

E=1000\times(0.10)^3\times4186\times41

E=171626\ J

We need to calculate the time

Using formula of time

t=\dfrac{E}{P}

Put the value into the formula

t=\dfrac{171626}{1285.02}

t=133.5\ sec

Hence, The time is 133.5 sec.

3 0
4 years ago
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.
7 0
3 years ago
"My partner seems to be more in the mood at night time, but I'm more in the mood in the morning. Why might that be and is there
Marianna [84]

Answer:

I don't get it?

like yhu want us to rate it or?

Explanation:

5 0
3 years ago
A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori
vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

  • The horizontal component of the pulling force, F cos \theta, where F = 50 N is the magnitude and \theta=60^{\circ} is the angle between the direction of the force and the horizontal; this force acts in the  forward direction
  • The force of friction, F_f, acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

\sum F_x = ma_x

where

m is the mass of the block

a_x is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

a_x = 0

So the equation becomes

\sum F_x = 0 (1)

The net force here is given by

\sum F_x = F cos \theta - F_f (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N

Learn more about  force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

4 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
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