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hram777 [196]
3 years ago
10

Volcanoes are a topic that would be studied in what branch of science

Physics
2 answers:
konstantin123 [22]3 years ago
7 0
The answer is A :) Hope this helps
Vitek1552 [10]3 years ago
5 0
Earth science, A would be the answer!
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What are the characteristics of a blackbody radiator? Give an example.
mihalych1998 [28]

Answer:

A blackbody, or Planckian radiator, is a cavity within a heated material from which heat cannot escape. No matter what the material, the walls of the cavity exhibit a characteristic spectral emission, which is a function of its temperature.

Example:

Emission from a blackbody is temperature dependent and at high temperature, a blackbody will emit a spectrum of photon energies that span the visible range, and therefore it will appear white. The Sun is an example of a high-temperature blackbody.

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3 years ago
Another name for the skin as a whole.
sergejj [24]
Organ, integument, dermis
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3 years ago
Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

f(x) = 8x^3-3x^2-5x^8

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0
4 years ago
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The energy emitted from the sun is a product of ________.
Tamiku [17]
Q. The energy emitted from the sun is a product of ________.

A. Fusion
3 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
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