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Oksanka [162]
2 years ago
12

Work and Power Practice Calculations

Physics
1 answer:
Elza [17]2 years ago
5 0

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

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What should not be used to support a scaffold because they could collapse, sending the structure to the ground?
uranmaximum [27]

Answer:

Cinder blocks

Explanation:

From safety training guides, whenever there is inadequate support, poor

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In light of this safety rule, Cinder blocks such as sandcrete blocks, bricks e.t.c shouldn't be used to support scaffold because they have a high tendency of being crushed under load

5 0
3 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
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grin007 [14]

Answer:

ecdbaf

Explanation:

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It is an extension of the grounding system
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This is because water evaperates after reaching it's boiling point.
4 0
3 years ago
Read 2 more answers
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