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vesna_86 [32]
3 years ago
12

A river flows at a velocity of 3 km/h relative to the riverbank. A boat moves downstream at a velocity of 15 km/h relative to th

e river. What is the velocity of the boat relative to the riverbank?
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
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Which method is not a technique currently used by ocean scientists to map the topography of the ocean floor?
raketka [301]

Answer:

Magnetometer

Explanation:

Magnetometer technique is not using by scientists for studying the ocean floor.The scientists currently is using SONAR ( sound navigation and ragging) technique for studying the ocean floor.SONAR is used sound waves sound waves for studying the ocean floor or we can say that SONAR is based on sound propagation.

Therefore answer is Magnetometer

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What is meant by the phrase " a consistent method of measurement "? someone help plss
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If volume is constant and heat is added, what happens to the
bezimeni [28]

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the temperature would Increase and pressure would increase

Explanation:

This would occur because the temperature would move to the liquid through conduction and the pressure would increase because the heat would cause more and more pressure

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Why isnt geothermal energy used more often​
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Risk of return on investment is higher than other forms of energy generation.
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Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s
Maru [420]

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

58 m/s -  9.8 m/s^2 *4.6 s = v0 = 12.92 m/s

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

y0 + v0*t + 1/2 g *t^2 = yt

replacing

0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m

5 0
3 years ago
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