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Gala2k [10]
3 years ago
13

Spaceships A and B are traveling directly toward each other at a speed 0.5c relative to the Earth, and each has a headlight aime

d toward the other ship.
What value do technicians on ship B get by measuring the speed of the light emitted by ship A's headlight?


Answer choices


1).75c


2)1.0c


3)1.5c


4) .5c
Physics
1 answer:
Rainbow [258]3 years ago
5 0
The answer is 2) 1.0c. Light will always propagate through a vacuum at the speed of light “c”; even when moving at a significant fraction of the speed of light, observers will still measure this as the speed of light and the difference is resultant of time dilation.
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Explanation:

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6 0
3 years ago
A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0
1 year ago
2 Points
lana66690 [7]

Answer:

the answer will be 3.763 seconds

6 0
3 years ago
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IrinaK [193]
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It's a number with a scalar value. It's a quantity with a vector.
It can't possibly be negative. It might be a negative number, a zero number, or a positive number.
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2 years ago
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