All categories

2 years ago

Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.

Physics

1 answer:

RSB [31]2 years ago

5 0

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

You might be interested in

Pls answerrrrr thisssss

Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0

3 years ago

An object at rest starts accelerating.

Shalnov [3]

Answer:

We are given:

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

Solving for 'a'

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0

3 years ago

Caroline conducts research on how the amount of fiber in a student's breakfast affects their grades in school. How should she re

Evgesh-ka [11]

Answer:

data table and line graph ( first choice)

8 0

3 years ago

Read 2 more answers

When gasoline is burned in a car engine,_____ energy is converted into _____ energy.

Kisachek [45]

Chemical Energy is converted into Mechanical Energy.

When gasoline is burned in a car engine, chemical energy is converted into mechanical energy.

8 0

3 years ago

Read 2 more answers

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp

Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_{i}$ = 60 mph = 26.8224 m/s

Final velocity $V_{f}$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $\frac{1}{2}$ m( $V_{i}$ ² - $V_{f}$ ² )

we substitute

Δk = $\frac{1}{2}$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $\frac{1}{2}$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0

3 years ago