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murzikaleks [220]
4 years ago
12

Question 25

Physics
1 answer:
kicyunya [14]4 years ago
8 0
The acceleration of gravity in low Earth orbit is a fraction smaller than it is on the Earth's surface, like maybe 15% less.

The astronauts feel no gravity at all, because they and everything around them are constantly falling.
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Ball rolls 6cm in one second. 6cm x 40 sec = 240cm
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Explain giving an example for;
Bingel [31]

\boxed{\large{\bold{\red{ANSWER~:) }}}}

  • The motion of a ball falling through the atmosphere, or a model rocket being launched up into the atmosphere are both excellent examples of Newton’s 1st law of motion.

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  • You hit a wall with a certain amount of force, and the wall returns that same amount of force. This is an example of Newton’s 3rd law of motion
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3 years ago
Who am I???????????????????????????????????
Tems11 [23]

Answer:

a human that walks on earth

Explanation:

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A block of mass 0.404 0.404 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring
VARVARA [1.3K]

To solve this problem it is necessary to apply the concepts related to the Force from Hook's law as well as the definition of the period provided by the same definition.

We know that the Force can be defined as

F = xk \rightarrow mg = kx \Rightarrow k = \frac{mg}{x}

Where

k = Spring constant

x = Displacement

g = Gravity

m = mass

At the same time the period of a spring mass system is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Where

m = Mass

k = Spring constant

Our values are given as,

m = 0.404kg

x = 0.666m

Replacing to find the value of the Spring constant we have that

k = \frac{mg}{x}

k = \frac{(0.404)(9.8)}{0.666}

k = 5.944N/m

Now using the formula of the period we know that

T = 2\pi \sqrt{\frac{m}{k}}

T = 2\pi \sqrt{\frac{0.404}{5.944}}

T = 1.638s

Finally, if the oscillation was 0.359m

The maximum height will be determined by the total length of that oscillation being equivalent to

h=2a

h = 2*0.359

h = 0.718m

4 0
3 years ago
A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is
Semenov [28]

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

5 0
3 years ago
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