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kenny6666 [7]
3 years ago
13

A mysterious device found in a forgotten laboratory accumulates charge at a rate specified by the expression q(t) = 9 − 10t C

from the moment it is switched on. (a) Calculate the total charge contained in the device at t = 0. (b) Calculate the total charge contained at t = 1 s. (c) Determine the current flowing into the device at t = 1 s, 3 s, and 10 s.
Physics
1 answer:
777dan777 [17]3 years ago
8 0

There is one mistake in the question is values of q(t) is not correctly written.The Correct question is here

A mysterious device found in a forgotten laboratory accumulates charge at a rate specified by the expression q(t) = 9−10t C from the moment it is switched on. (a) Calculate the total charge contained in the device at t = 0. (b) Calculate the total charge contained at t = 1 s. (c) Determine the current flowing into the device at t = 1 s, 3 s, and 10 s.

Answer:

(a) q(t)=9 C at t=0 s

(b) q(t)=-1 C at t=1 s

(c) I= -10 A at t=1 s , I= -10 A at t=3 s , I= -10 A at t=10 s

Explanation:

Given Data

q(t)=9-10t C

(a) t=0 s

(b) t=1 s

(c) t=1 s , t=3 s , t=10 s,

To Find

(a) q(t) at t=0 s

(b) q(t) at t=1 s

(c) I(current)

Solution

For part (a) of question

q(t)=9-10×t

at t=0 s

q(t)=9-10×(0)

q(t)= 9 C

For part (b) of question

q(t)=9-10×t

at t=1 s

q(t)=9-10×(1)

q(t)= -1 C

For part (c) of question

q(t)=9-10×t

as

I=\frac{dq}{dt}\\ I=\frac{d}{dt}(9-10t)\\ I=-10A

At t=1 s

I=-10 A

At t=3 s

I=-10 A

At=10 s

I=-10 A

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1) The landing spot of the projectile is given by d=v_x \sqrt{\frac{2h}{g}}

2) The common variable is the time

Explanation:

1)

The motion of a projectile consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

The landing spot can be determined in the following way:

- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation

h=ut+\frac{1}{2}at^2

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a=g=9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2h}{g}}

- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity v_x. Therefore, the horizontal distance covered is given by

d=v_x t

And substituting the time of flight,

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2)

Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by

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Learn more about projectile:

brainly.com/question/8751410

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