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3 years ago

Which of the following is a factor that determines the composition of magma?

Physics

1 answer:

Lina20 [59]3 years ago

5 0

The answer is C. I hope I helped!

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What is gravitational potential energy?

Kay [80]

Answer:

D). energy resulting from the attraction between two masses.

Explanation:

4 0

3 years ago

Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area

yaroslaw [1]

Answer:

$5.63\cdot 10^{-6} C$

Explanation:

The capacitor of a parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

A is the area of each plate

d is the separation between the plates

$\epsilon_0$ is the vacuum permittivity

The energy stored in a capacitor instead is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

$U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}$

And re-arranging it

$Q=\sqrt{\frac{2U\epsilon_0 A}{d}}$

Now if we substitute

$d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J$

We find the charge stored on the capacitor:

$Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C$

7 0

3 years ago

I need help fast!!!

Fantom [35]

You could easily help yourSELF if you simply follow the instructions in the question and "Use Ohm's law" to calculate the resistance.

Ohm's Law: Resistance = (voltage) / (current)

Resistance = (10 v) / (5 A)

Resistance = 2 ohms

8 0

3 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

Part A: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

Part B: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$