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vlada-n [284]
3 years ago
7

How does rusting differ from a combustion reaction

Physics
1 answer:
olga nikolaevna [1]3 years ago
7 0
Cool good luck with this 73737383+8282828=swo
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How do you get derived units from derived quantities
Tasya [4]

Answer:

A derived quantities is terms of the 7 base quantities via a system of quantity equations which are called SI derived units.

Explanation: there you go:)

4 0
3 years ago
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g
djyliett [7]

Answer:

a) m = 69.0 kg

b) release some gas in the opposite direction to the astronaut's movement

Explanation:

a) Let's use Newton's second law

         F = m a

         m = F / a

         m = 60.0 / 0.870

         m = 69.0 kg

b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved

         p₀ = p_f

         m v = M v '

         v ’= \frac{m}{M} \ v

so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.

One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.

3 0
3 years ago
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
When an object falls, it trades gravitational potential energy for kinetic energy, accelerating toward the ground. calculate the
liq [111]
The speed of the energy decreases
5 0
4 years ago
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