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3 years ago

What is the strength of the electric field 0.020 m from a 12 uc charge?

Physics

1 answer:

belka [17]3 years ago

5 0

Answer:

Option B is correct.

Explanation:

Given the charge, Q = 12uC

Now convert the uC into C = 5*10^-6 C

Given, d = 0.020 m

Now use the below formula:

E = K*Q/d^2

Now insert all the values in the above formula and then solve.

E = (9 ×10^9) × (12×10^-6) / (0.020)^2

E = 2.7 ×10^8 N/C

Therefore option B is correct.

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List two of the number of strategies you have learned

jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

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2 years ago

Why is it useful to calculate average speed?

Phantasy [73]

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3 years ago

A worker uses a cart to move a load of bricks weighing 680 N a distance of 10 m across a parking lot.

NNADVOKAT [17]

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3 years ago

Read 2 more answers

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago

(ASAP) would it be 125 m/s2 to calculate for her speeding up?

serg [7]

Answer:

$0\:\mathrm{ m/s^2}$

Explanation:

Recall the formula for acceleration:

$\displaystyle\\a=\frac{v_f-v_i}{\Delta t}$ , where $v_f$ is final velocity, $v_i$ is initial velocity, and $\Delta t$ is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

$\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}$

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0

2 years ago