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sweet [91]
3 years ago
13

Shruti is carrying around her 1.3kg laptop computer. What is the weight of this computer?

Physics
1 answer:
svet-max [94.6K]3 years ago
3 0

Answer:13newtons :w=mass by gravitation pull

Explanation:mass=1.3kg by (10n/kg)gravity on earth

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VERY URGENT GIVING BRAINLY​
lakkis [162]

Answer:

A

Explanation:

There is a mechanical advantage in this system....

    she only needed 100 N to lift 500 N

        1:5      she lifted it 2 meters, so if there WAS NO friction, she would have to pull in 5 x 2 = 10 meters

   but there IS friction so she must have pulled in MORE than 10 m  

8 0
2 years ago
Thanks to the development of transgenic plants, much of the food we eat has been altered so that it tastes better, looks better,
Sonja [21]
<span>Im guessing its genetics.??? </span>
5 0
3 years ago
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1. How much heat energy is required to raise the temperature of a 5 kg aluminium bar
Ray Of Light [21]

Answer:

180 kJ

Explanation:

Given that:

Mass (m) = 5 kg

Initial temperature (T1) = 28°C

Final temperature (T2) = 68°C

The change in temperature (ΔT) = T2 - T1 = 68°C - 28°C = 40°C

Specific heat capacity of aluminium (c) = 900 J/kg°C

The quantity of heat energy required (q) is given by:

q = mcΔT

q = 5 kg × 900 J/kg°C × 40°C

q =  180000 Joules

q = 180 kJ

Therefore 180 kJ is required to raise the temperature of aluminium from 28°C to 68°C.

5 0
3 years ago
Which workout schedule would be BEST suited for a weight-lifting workout that builds both strength and endurance
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My workout schedule is day on - day off

This helps the body rest after the work out and let it gain/build muscle mass

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4 0
4 years ago
Read 2 more answers
In each cycle, a heat engine an input of 1940 J of heat and exhausts 1480 J of heat. What is the thermal efficiency?
AleksAgata [21]

Answer:

0.237 (23.7 %)

Explanation:

The thermal efficiency of an engine is given by:

\eta=\frac{W}{Q_{in}}

where

W is the useful work output of the engine

Q_{in} is the heat in input

Here we have:

Q_{in}=1940 J

and the work done is the total heat in input minus the heat exhausted:

W=1940 J - 1480 J=460 J

So, the efficiency is

\eta=\frac{460 J}{1940 J}=0.237 (23.7 %)

8 0
3 years ago
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