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3 years ago

A light-year is a measurement of _____. a. time b. gravity c. energy d. distance

Physics

2 answers:

Evgen [1.6K]3 years ago

6 0

D. distance

A light-year is the distance light would travel in 1 year.

Paladinen [302]3 years ago

3 0

the answer to your question is D .Hoped this helps

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A centrifuge in a medical laboratory rotates at an

suter [353]

Answer:

- 273.77 rad/s^2

Explanation:

fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps

f = 0

ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s

ω = 2 π f = 0

θ = 46 revolutions = 46 x 2π radian = 288.88 radian

Let α be the angular acceleration of the centrifuge

Use third equation of motion for rotational motion

$\omega^{2}=\omega _{0}^{2}+2\alpha \theta$

$0^{2}=397.71^{2}+2 \times \alpha \times 288.88$

α = - 273.77 rad/s^2

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3 years ago

Vector A and B are given as follows:

Goshia [24]

Answers

2i+3j+4k

Explanation:

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3 years ago

A cosmic catastrophic event occurred that caused the tilt of the Earth's axis relative to its plane of orbit to increase from 23

Gnom [1K]

Answer: The elimination of seasonal variations

Explanation:

Since the cosmic catastrophic event which occurred led to the tilt of the Earth's axis relative to the plane of orbit to increase from 23.5° to 90°, the most obvious effect of this change would be the elimination of seasonal variations.

It should be noted that seasonal variation refers to the variation in a time series that's within a year which is repeated. The cause of seasonal variation can include rainfall, temperature, etc.

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3 years ago

Classify the following situations as involving balanced or unbalanced forces.

Murrr4er [49]

Answer:

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Explanation:

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5 0

3 years ago

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$ ,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0

3 years ago