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s2008m [1.1K]
3 years ago
10

Conversion fraction 1$=4q, how many are in 20$

Physics
2 answers:
Tom [10]3 years ago
8 0

Answer:

\boxed{\sf 20 \$ = 80q}

Given:

1$ = 4q

To Find:

How many quarters are in 20$

Explanation:

To find out how many quarters are in 20$ we need to multiple 4 × 20.

\sf 1\$  = 4q

\sf  20\$  = 4 \times 20q

\sf = 80q

Over [174]3 years ago
7 0

Answer:

\boxed{\sf 20 \$ = 80q}

Given:

1$ = 4q

To Find:

How many quarters are in 20$

Explanation:

To find out how many quarters are in 20$ we need to multiple 4 × 20.

\sf 1\$  = 4q

\sf  20\$  = 4 \times 20q

\sf = 80q

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A scuba tank, when fully submerged, displaces 16.7 L of seawater. The tank itself has a mass of 13.4 kg and, when "full," contai
meriva

Answer:

m = 17.87 kg

Explanation:

As we know that tank itself has mass given as 13.4 kg

so it is given as

m_1 = 13.4 kg

also the tank contains air in it and the mass of air inside the tank is also given as 4.47 kg

so it is

m_2 = 4.47 kg

so total mass of the tank is the mass of the empty tank and mass of air in the tank

m = m_1 + m_2

m = 13.4 + 4.47

m = 17.87 kg

3 0
3 years ago
Match each situation below to a letter on the illustration.
kari74 [83]

Answer:

What illustration

Explanation:

7 0
3 years ago
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:

a)

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

b)

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.

c)

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.

d)

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.

e)

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

So

(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)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
A heavy box is in the back of a truck. the truck is accelerating to the right. apply your analysis to the box.a) draw a motion d
Andru [333]

Explanation :

It is given that, a heavy box is in the back of a truck and the truck is accelerating to the right.

The description is shown in<em> figure 1</em>. Here, the forces that act on both box and the truck are the normal force, the frictional force and their weight mg.

The free body diagram is shown in <em>figure 2</em>. This shows the total forces            ( F = ma )acting on the truck. As a result, the truck will move in a forward direction.

Hence, this is the required explanation.

5 0
3 years ago
Read 2 more answers
What makes up a nucleus of an atom
Tanzania [10]
The nucleus is the center of an atom.
6 0
4 years ago
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