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Radda [10]
2 years ago
9

Given: Q1 = +10 uc = 1.0 x 10-5C

Physics
1 answer:
ser-zykov [4K]2 years ago
5 0

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

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2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal
Aleksandr [31]

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is

<em>x</em> = (7.2 m/s) <em>t</em>

The object's height <em>y</em> at time <em>t</em> is

<em>y</em> = 9.4 m - 1/2 <em>gt</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

<em>v</em> = -<em>gt</em>

(a) The object hits the ground when <em>y</em> = 0:

0 = 9.4 m - 1/2 <em>gt</em>²

<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)

<em>t</em> ≈ 1.92 s

at which time the object's vertical velocity is

<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

8 0
2 years ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0
2 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 81 mN when sepa
Gekata [30.6K]

Answer:

Explanation:

Check the attachment for solution

8 0
3 years ago
What is the function of the electron transport chain in cellular respiration. why is oxygen needed for the electron transport ch
SpyIntel [72]
1) <span>The function of the electron transport chain is to pump protons in the mitochondrion inter-membrane, thus building up a proton gradient. This gradient will allow the ATP syntheses</span><span>.</span>

2) Why we need oxygen for the electron transport chain:
 At the end of the electron transport chain is the Oxygen that will accept electrons and picks up protons to form water. If the oxygen molecule is not there the electron transport chain will stop running, and ATP will no longer be produced. Basically, we need the oxygen to produce more ATP.


8 0
3 years ago
A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie
Harman [31]

Answer:

Explanation:

Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².

Now, the pilot is traveling in a circle of radius

r = 3340 m

And the speed is

v = 495 m/s

Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

a = v² / r

a = 495² / 3340

a = 73.36 m/s².

Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully

4 0
3 years ago
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