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BaLLatris [955]
3 years ago
15

A book with a mass of 1kg is dropped from a height of 3m . What is the potential energy of th book when it reaches the floor?​

Physics
1 answer:
bonufazy [111]3 years ago
4 0

Answer:

30 J

Explanation:

height = 3 m

mass = 1kg

acceleration due to gravity = 10m/s²

Potential Energy = mass× gravity × height

= 1kg ×10m/s²× 3m

= 30 J

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How does the density of water change when: (a) it is heated from 0o
Radda [10]

Answer:

[b] it id heated from 4o

Explanation:

3 0
3 years ago
a force of 25 newtons moves a box a distance of 4 meeters in 5 seconds.the work done on the box is ? NM and the power is. ? nm/
miskamm [114]

Answer:

The work done on the box is 100 Nm

The power is 20 Nm/s

Explanation:

There is a force 25 newtons moves a box a distance of 4 meters in

5 seconds

The work done on the box is the product of the force and the distance

that the box moves ⇒ <em>work = force × distance</em>

The force = 25 newtons

the distance = 4 meters

Work = 25 × 4 = 100 NM

<em>The work done on the box is 100 Nm</em>

<em></em>

The force moves the box 4 meters in 5 seconds

The power is the rate of work

<em>The power = work ÷ time</em>

The work = 100 Nm

The time = 5 seconds

The power = 100 ÷ 5 = 20 Nm/s

<em>The power is 20 Nm/s</em>

6 0
3 years ago
Can someone explain to how to calculate this
Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

3 0
3 years ago
Which sphere of Earth do humans exist in?
PolarNik [594]
Biosphere is the answer, hope i helped :))
4 0
2 years ago
Read 2 more answers
An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra
slega [8]

Answer:

Part a)

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

E = 4.4 \times 10^{-13} J

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

m_1v_1 + m_2v_2 + m_3v_3 = 0

(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0

(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

By equation of kinetic energy we have

E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2

E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}

E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}

E = 4.4 \times 10^{-13} J

8 0
3 years ago
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