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vagabundo [1.1K]
3 years ago
7

The blackbody curve of a star moving toward Earth would have its peak shifted (a) to lower intensity; (b) toward higher energies

; (c) toward longer wavelengths; (d) toward lower energies.

Physics
1 answer:
trasher [3.6K]3 years ago
8 0

As show the figure, peak of each blackbody curves shifts towards higher frequency. Then energy emitted by the black body is directly proportional to the frequency incident radiation.

Hence peak is shifted to higher energies.

Therefore the correct option is B: Toward higher energies.

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Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha
Ann [662]

Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

8 0
3 years ago
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of which fun
gizmo_the_mogwai [7]

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

<h3>What is vertical motion of a projecile?</h3>

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

<h3>What is horizontal motion of a projecile?</h3>

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

<h3>Direction of the motion</h3>

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

Learn more about vertical motion here: brainly.com/question/24216590

#SPJ4

4 0
2 years ago
A ball is thrown vertically upwards with a velocity
zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

6 0
3 years ago
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0
3 years ago
A 2,000 kg car travels with a tangential
Gekata [30.6K]
A= v²/R
a = 12²/30 =4.8 m/s²
3 0
3 years ago
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