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Dafna11 [192]
2 years ago
10

WILL GIVE BRAINLIEST PLEASE ANSWER ALL QUESTIONS

Physics
1 answer:
Rainbow [258]2 years ago
5 0
There would be more - charges on the peanuts because they gain electrons which are represented with the - sign. There would also be more + on the cat to show that the cat’s fur is positively charged. The peanuts would be attracted to the cat because opposite charges attract.

As for explaining the first model, there should be the same amount of + and - charges on the peanuts as well as the cat to show that it is balanced and the peanuts are not attracted or repelled by the cat.
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PLEASE HURRYYY
babymother [125]

your answer is make up artist

4 0
4 years ago
75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0
3 years ago
A wall (of thermal conductivity 0.98 W/m · ◦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside
disa [49]

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

4 0
2 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0
4 years ago
Can someone please help with this. "A ball is thrown vertically upward with a speed of 26.6 m/s. How high does it rise? The acce
lubasha [3.4K]

v^2=u^2+2as

0=26.6^2-2x9.8xs

-26.6^2/-2x9.8=s

calculator

8 0
3 years ago
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