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katrin2010 [14]
3 years ago
15

(15 points) :^|

Physics
1 answer:
mihalych1998 [28]3 years ago
7 0
That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
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PLS HELP ASAP (no links)
nikklg [1K]

Answer:

The mass goes down

Explanation:

Because mass is the quantity of matter contained in a substance And Volume is the space occupied by a substance. So when the volume is less the mass decreases

6 0
2 years ago
Read 2 more answers
Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu
morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

U=mgh

U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0
3 years ago
Simplify 6.25 − 8.<br><br>please help
tatyana61 [14]

6.25 - 8 = -1.75

Hope this helps

-AaronWiseIsBae

7 0
2 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
if the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
diamong [38]
So base on your question that as if the vapors volume were to incorrectly recorded as 125ml, the effect of the error to calculate the molar mass is the same as the error in measuring the volume of the vapor. I hope you are satisfied with my answer and feel free to ask for more 
7 0
3 years ago
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