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2 years ago

Which change increases the electric force between objects?

Physics

2 answers:

sdas [7]2 years ago

7 0

the answer to your question is b broskey

Brrunno [24]2 years ago

3 0

Well, the force is proportional to the product of the charges
on the two objects. So if the objects are already negatively
charged distance between them is unchanged, then adding
electrons to either or both objects would increase the forces
between them.

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The humidity you feel in the southern states of the U.S. is _____. part of the water cycle water vapor caused by the Sun's heat

Likurg_2 [28]

Hey there!

The answer is :
The humidity you feel in the southern states of the U.S. is water vapor
water vapor is between a gas and liquid so that is what gives a humid feel in the hot sun

Hopes this helps u :D

4 0

2 years ago

Which energy sources can transfer thermal energy to a heat pump? Check all that apply.

choli [55]

Answer:1.Air, 2.the ground,3.water

Explanation:I just got it right:)

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3 years ago

Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum

loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

$R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45$

The minimum sum of the resultant vector,

$R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10$

So, the magnitude of this sum either 45 or -10.

6 0

2 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

6 0

3 years ago

The net force acting on the ball below is ___ N

yKpoI14uk [10]

The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).

8 0

2 years ago