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frosja888 [35]
3 years ago
12

Which of the following is true about the Earth's natural resources?

Physics
1 answer:
Art [367]3 years ago
4 0
The last answer is the only correct one. Do you want to know why or were you just checking?
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A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th
mr Goodwill [35]

elasticity stretches and can also return to it's normal size ..

8 0
2 years ago
A rectangular box has side 10cm x 5cm x 2cm. If it lies on a horizontal surface and has weight 1.00N, calculate the maximum and
Vladimir [108]

Answer:

P max = 1000 pa

P min = 200 pa

Explanation:

P = F/A

pressure will be maximum when surface gets minimum. so to find the maximum amount of pressure we need to calculate the minimum surface. it is 2cm×5cm = 10cm² = 0.001m² . then we have:

P = 1 / 0.001 = 1000 pa

to find minimum pressure the surface that must be chosen is 10cm×5cm = 50cm² = 0.005m² .

P = 1 / 0.005 = 200 pa

7 0
2 years ago
A calcium bromide compound is represented by which formula?
ipn [44]

Answer:

Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.

4 0
3 years ago
Read 2 more answers
Please need help on this
REY [17]
The handle of a metal pot gets warm when the water inside the pot starts to boil
8 0
3 years ago
This procedure has been used to "weigh" astronauts in space: A 42.5 kg chair is attached to a spring and allowed to oscillate. W
Bingel [31]

Answer:

The mass of the astronaut is approximately 119.74 kg

Explanation:

Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:

T=2\pi\sqrt{\frac{m}{k}} (1)

First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:

<em>(it’s important to note that one complete vibration is the period of the movement)</em>

1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2}

k\approx992.8\,\frac{N}{m}

Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds

2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5

\mathbf{M_{a}\approx119.74\,kg}

3 0
3 years ago
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