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3 years ago

If the number of particles of a gas is increased, the gas pressure will do what?

Physics

1 answer:

Inessa [10]3 years ago

5 0

Gas pressure will also increase then.., if volume is kept constant

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In what direction would you head from New York state to find warmer temperatures

Anni [7]

Head south. If you traveled west, it would gradually get warmer because the Pacific Coast is warmer than the Atlantic. But the quickest change would be if you traveled south to North Carolina and there you would experience warmer temperatures.
The climate would still be humid as it's still near the coast, but since it's closer to the equator, it would be warmer.

3 0

3 years ago

A boat heads north at 30 mph while a steamer heads east at 40 mph at the same time. When is the distance between them 50 mi? Ans

Reil [10]

Answer:

1 hour

Explanation:

Speed of the first boat = 30 mph

Speed of the second = 40 mph

The boats will cover different distances but the time taken will be the same.

Time taken by the boats = t

Distance = Speed × Time

Distance covered by the first boat = 30t

Distance covered by the second boat = 40t

Distance between the boats = 50 mi

From the Pythagoras theorem

$\sqrt{(30t)^2+(40t)^2}=50\\\Rightarrow \sqrt{2500t^2}=50\\\Rightarrow 50t=50\\\Rightarrow t=\frac{50}{50}=1\ hour$

Time taken by the boats when they are 50 mi away is 1 hour

7 0

3 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

Which of the view will show you a view<br>to the<br>very<br>Similar<br>Print View?

sweet [91]

What are you talking about

3 0

3 years ago

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$