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alukav5142 [94]
3 years ago
9

If a star’s radiation peaks in the ultraviolet region of the spectrum, what could you conclude?

Physics
2 answers:
Alex_Xolod [135]3 years ago
4 0

Answer:

It is a very intense star

Explanation:

As we know by Electromagnetic radiations the energy of spectrum in decreasing order is given as

1) Gamma Rays

2) X rays

3) Ultraviolet Rays

4) Visible Light

5) Infrared waves

6) Microwaves

7) Radio waves

So here we can say that ultraviolet rays are at peak intensity for the spectrum in the radiation which means major proportion of the spectrum is consisting this part of the electromagnetic waves

So here the star would be very intense due to large number of this spectrum.

PSYCHO15rus [73]3 years ago
3 0
The short answer is "it is a very intense star." 
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An object experiences an impulse, moves and attains a momentum of 200 kg·m/s. If its mass is 50 kg, what is its velocity?
Nady [450]

Answer:

4 m/s

Explanation:

Momentum is defined as:

p=mv

where

m is the mass of the object

v is its velocity

For the object in this problem, we know:

p = 200 kg m/s is the momentum

m = 50 kg is the mass

Solving for the velocity, we find:

v=\frac{p}{m}=\frac{200}{50}=4 m/s

8 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
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
Oliga [24]

Answer:

The value of change in internal l energy of the gas = 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

6 0
3 years ago
In 1 to 2 sentences explain how you would determine the constant K for a spring
Olin [163]

When spring times comes around you ever be like, "k." If you feel like that every year during spring then there you go

8 0
3 years ago
Two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of
olga2289 [7]

Answer:

c. because A will land first becuase its heavier :)

Explanation:

8 0
2 years ago
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