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3 years ago

1. Line segment AC touches the circle at a single point B. Line segment OB extends through the center of the circle.

Physics

1 answer:

adelina 88 [10]3 years ago

8 0

AC is a tangent to the circle

A tangent is perpendicular to the diameter of the circle, so B is 90 degrees

Radius is half the diameter, so 3.5*2 = 7m

Area = (pi)(r^2), so (pi)*3.5^2 = 38.48m^2

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A body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency o

Scilla [17]

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

$\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s$

(b) The frequency of oscillation is given by :

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz$

Hence, this is the required solution.

7 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

What is not changed when work is done by a machine?

den301095 [7]

B) The amount of work done

4 0

3 years ago

Read 2 more answers

Consider dropping a ball from rest. This ball moves from astate of high gravitational potential energy to one of lowgravitationa

Leni [432]

Answer:

From the negative to the positive cable.

Explanation:

The electrons have negative charge, which means that the negative terminal of the battery will suply the electrons, thus they are present in excess on the negative cable and will jump from it to the positive cable. This current direction is called real current.

3 0

3 years ago

In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.

garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

$\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...$