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2 years ago

2 Pont

Physics

1 answer:

Reptile [31]2 years ago

6 0

Answer:

A. It makes astronauts weightless.

Explanation:

Gravity does not make astronauts feel weightless. Astronauts are weightless because they are orbiting at the same rate as their shuttle.

Although the force of gravity weakens as one moves away from the earth surface, it does not mean that this force is absent in orbit

Gravitational force has a constant acceleration value near the earth surface which is commonly known to be 9.8m/s².
It is a force of attraction tending to hold and bind bodies together so far they have mass.
This force keeps every thing from escaping space-ward from the earth surface.

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Find the magnitude of this vector: 174 m N 188.4 m HELP FAST

Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

$a^2+b^2=c^2$

Plug values in

$88.4^2+174^2=c^2$

Simplify

$7814.56+30276=c^2$

Add the two values

$38090.56=c^2$

Take the square root of both sides

$195.168\approx195.168$

8 0

2 years ago

Early psychologists were predominately all of the following except __________. A. white B. male C. European D. wealthy

yulyashka [42]

Answer: D. wealthy

Explanation: on the e2020 test its right

7 0

3 years ago

Read 2 more answers

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

(10p+15,15-10q)=(25,5)

Harrizon [31]

63783626736377474737377447

8 0

2 years ago

Which occurs at a transform boundary?

andreev551 [17]

The answer is B. One plate slides past another.

The San Andreas Fault in California and the Alpine Fault in New Zealand are examples of transform boundaries.

Hope this helps! :)

5 0

2 years ago