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Art [367]
2 years ago
12

Who was known for being a pilot and an astronaut that walked on the moon?.

Physics
2 answers:
vodka [1.7K]2 years ago
4 0

Answer:

Neil Armstrong was the astronaut

Mkey [24]2 years ago
3 0

Answer:

albert einsteinssssssssssssssssssssssssss

Explanation:

You might be interested in
If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?
777dan777 [17]

Answer:

It’s 18.0 m/s

Explanation:

Use acceleration formula then plug in 9.8 and 1.84s

6 0
3 years ago
An electron is released form rest in a region of space where a uniform electric field is present. Joanna claims that its kinetic
djyliett [7]

Answer:

Neither.

Explanation:

When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).

It will move towards a point  with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:

⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)

As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.

4 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
2) A motorcycle is moving at a constant speed of 40 km/h. How long does it take the motorcycle to
docker41 [41]

Answer:

2 hours

Explanation:

The motorcycle travels 40 km per hour.

80km / 40km/h = 2 hours.

7 0
4 years ago
Read 2 more answers
A student environmental group is creating a campaign for locally sourced energy resources.What would be best for them to feature
Ivenika [448]

Answer:

<em>A farmer with a field of solar panels.</em>

Explanation:

The closest to a locally sources energy would have been

A coal mine located in their county.

But coal as an energy source is not environmentally friendly due to carbon emission, and should not be what the group should advocate for.

<em>The best bet for them is </em>

<em>A farmer with a field of solar panels.</em>

As solar panels are a source of green energy and green energy is what the environmental group should often and always advocate for

3 0
3 years ago
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