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Lisa [10]
3 years ago
5

Which one is not a derived unit?1. Hertz2. mol3. Watt4. Newton​

Physics
2 answers:
Nesterboy [21]3 years ago
6 0
The mol is not a derived unit.
elena55 [62]3 years ago
5 0

Answer:

mol

Explanation:

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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
A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh
melisa1 [442]

Answer: 1.2m/s^2

Explanation: the force exerted on the car is 900N upwards

The mass of the car is 750kg

According to Newton's third law acceleration is proportional to force

F = ma

900 = 750a

a = 900/750

a = 1.2m/s^2

5 0
4 years ago
A baby is dropped from a 30 m cliff on Earth. Acceleration due to gravity on Earth is about 9.8 m/s^2. Calculate the following q
Katarina [22]

Answer:Dropping is not a good predictor of when labor will begin. In first-time mothers, dropping usually occurs 2 to 4 weeks before delivery, but it can happen earlier. In women who have already had children, the baby may not drop until labor begins.

Explanation:

4 0
3 years ago
???????????!??!!!!!!!!​
Diano4ka-milaya [45]
The answer would be B.

6 0
3 years ago
Calculate the pressure exerted by a man having mass 50 kg standing on one feet having area 500cm2​
Burka [1]

Answer:

9800 Pa

Explanation:

Data: pressure-?

Force- mass ×acceleration

= 50× 9.8m/s^2

=490N

Area= 500cm^2=0.05m^2

pressure= force/ area

p=F/A

P= (490N)/( 0.05m^2)

=9800 Pa

7 0
3 years ago
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