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olasank [31]
3 years ago
13

How far above the surface of earth would you weigh 71.5 percent of your surface weight?

Physics
1 answer:
kramer3 years ago
4 0
Hey, Name's Jessy. I hope, I answer your question.

Nearly 400 lbs,by 11.73%.
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It's five miles from Tim's house to Rita's house. Roughly how long is this distance in feet?
sergey [27]

Answer:

26400 ft

Explanation:

that would be my answer

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3 years ago
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What are the conditions necessary for a terrestrial planet to have a strong magnetic field?.
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Both a molten metallic core and reasonably fast rotation.
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2 years ago
If an antelope can run 14.1 km in 39 min, how fast is it km/h(for reference , 110 km/h is 65 mph )
shtirl [24]
It might make more sense putting it another way but this is basically it. you just take the minutes and divide them by 60 to convert them to hours. then simplify the ratio

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3 years ago
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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J
olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0
3 years ago
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