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Westkost [7]
3 years ago
13

Match the type of metamorphism to its best definition.

Physics
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

See Explanation.

Explanation:

Impact metamorphism:

The process in which a rock changes because of extreme heat and pressure caused by a meteorite collision with Earth.

Regional metamorphism:

The process in which buried rocks in a large area are changed by heat, pressure, and fluids.

Contact metamorphism:

The process in which a rock changes as a result

of heat generated from magma.

Dynamic metamorphism :

The process in which a rock changes due to

pressure directed on it.

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Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
A negative ion is (larger/smaller) than its parent atom
Lerok [7]
Its larger and if u where wondering to positive ions are smaller
6 0
3 years ago
How does the latent heat of fusion of water help slow the decrease in air temperature,perhaps preventing temperatures from falli
love history [14]

Answer:

Water has the ability to release a large amount of energy during the freezing process. All of this energy is released into the air, resulting in a greater movement of air particles that will increase the air temperature. This heat will prevent air temperatures from drastically falling below 0ºC.

Explanation:

The latent heat of water occurs when the water is changing its physical state. In other words, when substances are changing their physical state, the amount of heat calculated for this change is called latent heat, and as we have already said, this heat is not related to the heat exchanges between two systems, but the change of state. physicist.

When water is changing from its liquid state to a solid state, we call latent heat latent heat from fusion (this is because fusion is the name we give to when liquid water is turning to ice). In the process of freezing the page, latent heat releases a large amount of energy into the air near the water. This energy agitates the air molecules, generating heat and preventing the air from reaching temperatures below 0ºC.

6 0
3 years ago
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
JulijaS [17]

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

5 0
3 years ago
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
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