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2 years ago

The graph below shows the blackbody radiation curves for four stars.

Physics

1 answer:

8090 [49]2 years ago

7 0

The right answer for the question that is being asked and shown above is that: "The star with a wavelength of approximately 500 nm does not emit light in the red part of the spectrum.<span>" Based on the graph, wavelengths at approximately 500mn has less chance or can not emit light at the red part.</span>

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A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1

const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0

2 years ago

Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an

gulaghasi [49]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct answer is option c

Explanation:

Faraday states that when there is a change in magnetic field of a coil of a wire, it means that there exist an emf in the circuit which in induced due to the change in the magnetic flux

From the question two separate but nearby coils are mounted along the axis. First coil is connected to the power supply and the current flow is controlled by the supply.When the current alternates, it would produce magnetic field ,also the second coil is connected to an ammeter which indicates the current that is flowing in it when current in the first coil changes

This magnetic field that is produce would cause a change flux which would induce current in the second coil so the ammeter would indicate current flow in the second coil

a is incorrect because the current in fir coil is not change hence flux won't change therefore current is is not induced in second coil

This is the same reason b is incorrect

d is incorrect due to the fact that when the second coil is connected to a power supply by rewiring it to be in series with first coil the law of electromagnetism would no longer hold so he ammeter would show no reading

6 0

2 years ago

Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13

Julli [10]

The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s

3 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Basile [38]

The strength of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 is the number of turns in the coil

$\Delta \Phi$ is the change in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ is the time interval

$\epsilon = 0.166 V$

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

$\Delta \Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area of the coil

$\theta=0^{\circ}$ is the angle between the normal to the coil and the field

The area of the coil can be written as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ is its radius

Substituting everything into eq.(1) and solving for B, we find:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

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8 0

2 years ago

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