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3 years ago

An isolated conductor has a net charge of +12.0 × 10- 6 c and a cavity with a particle of charge q = +3.70 × 10-6

1 answer:

8 0

This is just a simple problem finding out the outer surface charge, the inner surface charge and the net charge. Net charge by definition means the difference between two charges. In this case, the formula that is applicable here is outer surface charge = total net charge - inner cavity surface charge. Since we are given already with the net charge equal to 12.0 x10-6 C and the inner charge magnituude f 3.7 x10-6 C, the the total charge must be outer charge is +10x10(-6)) - (-3.0x10(-6)) = +1.3x10(-5) C.
Charges are measured in coloumbs and most likely exist on surfaces of entities like particles, walls etc.

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Which of the following is not a way to make exercising on a cold day safer?

IRISSAK [1]

Answer:

Explanation:

I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)

4 0

3 years ago

Read 2 more answers

A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

koban [17]

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

3 0

3 years ago

An object travels with a constant speed in a circular path. The net force on the object is

Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

$F=ma$

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

8 0

3 years ago

Two of the three tuning forks have known frequencies. When a 610 Hz fork and the unknown fork are struck together, four beats pe

RUDIKE [14]

-- In combination with 610 Hz, the beat frequency is 4 Hz.

So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.

In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.

-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.

So the unknown is (605+9) = 614 Hz.

6 0

4 years ago

A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the a

Verizon [17]

Answer:

a) The average velocity is 32 m/s

b) See the attached figure. Slope of the line = 32.

Graphic: a line that passes through the points (0; 50) and (5; 210)

Explanation:

a) The average velocity is calculated as the displacement over time:

v = ΔX / Δt

where

ΔX : displacement ( final position - initial position)

Δt : time (final time - initial time)

Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

Δt = 5 s

v = 160 m / 5 s = 32 m/s

b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

4 0

3 years ago