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Sedbober [7]
3 years ago
6

Water is traveling into what sphere during infiltration?

Physics
1 answer:
o-na [289]3 years ago
5 0
When precipitation hits the ground and hydrosphere
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What happens in a tug of war if the net forces are balanced and why?
FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

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Using clay electric rates, how much does it cost to watch Netflix for 4 hours on a smart tv?
OverLord2011 [107]

Answer: TVs cost between $0.0015 and $0.0176

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2 years ago
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0
3 years ago
You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica
Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A} , where:

l is the length

m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3

We divide the first equation by the second equation to get:

\frac{m}{R} = \frac{d A^2}{\rho}

A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8}   \ m^2

Using this Area, we find the diameter of the wire:

D = \sqrt{\frac{4A}{\pi}}

D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}

D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm

To find the length, we multiply the two equations stated initially:

mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}

l^2 = 8.534\\l =   2.92 \ m

8 0
3 years ago
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