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NISA [10]
3 years ago
15

How are energy and distance up the ramp related

Physics
1 answer:
AfilCa [17]3 years ago
8 0
I don't exactly understand the question but I'll try my best to answer it and help you. You need energy in order to go up the ramp and cover the distance. I hope this helps! :)
You might be interested in
How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?
Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 0
3 years ago
Which of the following is quantitative data? a. Color c. Shape b. Odor d. Volume
Papessa [141]
Hello there! Quantitive data has to do with measurements that can be shown with numbers. Examples of this are things like your height and the length of your arms. With that alone, A and B are eliminated, because those answer choices make no sense. They can't be expressed by numbers and you can't measure colors or odors mathematically. Volume is a way to measure something that CAN be written down by numbers. D is the only answer choice that fits the definition of quantitive data. The answer is D: volume.
6 0
3 years ago
) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene
Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
A student is helping her teacher move a 9.5 kg box of books. What net sideways force must she exert on the box to slide it acros
melisa1 [442]
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10-27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
8 0
3 years ago
Read 2 more answers
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