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Dennis_Churaev [7]
2 years ago
6

A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.

Physics
1 answer:
dlinn [17]2 years ago
5 0

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

  • Force= 9.80665 x 0.5kg
  • Force= 4.90332 Newtons

  • Distance changed= 5-3
  • distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

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Spidermans nemesis electro delivers 4kj of electrical energy in half a second how much power does it draw from the mains?
Elodia [21]

Power delivered = (energy delivered) / (time to deliver the energy)

Power delivered = (4,000 J) / (0.5 sec)

Power delivered = 8,000 watts

I'm a little surprised to learn that Electro draws his power from the mains.  This is VERY good news for Spiderman !  It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.

But OK.  Let's assume that Electro draws it all from the mains.  Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).

The efficiency of Electro's internal process is

<em>(power he shoots out) / (power he draws from the mains) </em>.

So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of

<em>(8,000 watts) / (his internal efficiency) .  </em>

4 0
3 years ago
Read 2 more answers
1) Describe how the atoms of a solid differ from the atoms of a liquid.
lisabon 2012 [21]

Answer:

liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

6 0
3 years ago
Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the m
AfilCa [17]

Answer: length of B =4.00

Explanation:

for  the vectors A and B and the angle between them as  x.

Magnitude of the sum of A and B is  given as = √(A²+B²+2ABcosx

where

Magnitude of A  = 3.00

Magnitude of the sum of A and B is  5.00

5.00=√(A²+B²+2ABcos90°

5.00= √3² +b² +0

5²= 3² +b²

25=9+b²

b²= 25-9

b² = 16

b=  √16

b= 4

7 0
3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
A concave lens has a focal length of 20 cm. A real object is 30 cm from the lens. Where is the image? What is the magnification?
Pani-rosa [81]

Answer:

12 cm and 0.4

Explanation:

f = - 20 cm, u = - 30 cm

Let v be the position of image and m be the magnification.

Use lens equation

1 / f = 1 / v - 1 / u

- 1 / 20 = 1 / v + 1 / 30

1 / v = - 5 / 60

v = - 12 cm

m = v / u = - 12 / (-30) = 0.4

6 0
3 years ago
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