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4 years ago

The continental crust has the average composition of _____.

Physics

2 answers:

Aliun [14]4 years ago

8 0

GRANITE is the correct answer!!:)

Hope this helps plz mark as brainlist and 5 star it would mean alot!!

kramer4 years ago

3 0

The answer would be, "Granite".

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What is it called when a solid goes to a gas?

murzikaleks [220]

Answer:

sublimation

Explanation:

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3 years ago

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If you add together all of the forces exerted on a object and get a non zero value that is called

True [87]

Hello There!

If you add together all of the forces exerted on a object and get a non zero value that is called the "NET FORCE" of the object

3 0

3 years ago

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A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s

Bogdan [553]

Answer:

a) $v=20.3m/s$

b) $a=2.35m/s^2$

Explanation:

From the exercise we know:

$x_{1}=30m$ ; $v_{1}=10m/s$ ; $t_{1}=3s$

$x_{2}=375m$ ; $v_{2}=50m/s$ ; $t_{2}=20s$

The formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

a) $v=\frac{375m-30m}{20s-3s}=20.3m/s$

The formula for average acceleration is:

$a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

b) $a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2$

4 0

3 years ago

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

KIM [24]

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

6 0

3 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago