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3 years ago

When converted to a household measurement, 9 kilograms is approximately equal to a

Physics

1 answer:

mariarad [96]3 years ago

3 0

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

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Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

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6 0

3 years ago

Read 2 more answers

URGENT!! ILL GIVE BRAINLIEST! AND 100 POINTS

lubasha [3.4K]

The inertia with the speed of 2.9 m/s has moved a distance of 3.6 kilometers in the west direction and 2.35 kilometers in the north direction.

<h3>What are trigonometric ratio?</h3>

In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others.

Given:

Speed of the speed 2.9 m/s

The direction in which the inertia change is moving = 43° north west.

Time taken to travel the distance = 30 minutes

According to the question that is moving in a Northwest direction with a velocity of 2.9 m per second.

If you want to find the distance it has moved in either North or west direction you will first have to find the total distance that kayak has moved in 30 minutes.

The distance travel by the kayak in 30 minutes = speed of person × time taken

The distance travel by the person in 30 minutes = 2.9 × 30 × 60

The distance travel by the person in 30 minutes = 5220 metre

The distance travel by the person in 30 minutes = 5.22 km

The distance traveled in the west direction can be find using trigonometric ratio.

5.2 × sin 43 = The distance traveled in the west direction

The distance traveled in the west direction = 3.6 kilometers

The distance traveled in the west = cos 43 × 5.2

The distance traveled in the west = 2.35 km

Therefore, The person with the speed of 2.9 m/s has moved a distance of 3.6 kilometers in the west direction and 2.35 kilometers in the north direction.

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6 0

2 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

$t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s$

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

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When heat is added to a substance, describe how the molecules are affected. Use the words kinetic energy and potential energy ap

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