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mariarad [96]
3 years ago
14

The quantity of charge passing through a surface of area 1.82 cm2 varies with time as q = q1 t 3 + q2 t + q3 , where q1 = 5.2 C/

s 3 , q2 = 2.5 C/s, q3 = 6.5 C, and t is in seconds. What is the instantaneous current through the surface at t = 1.1 s? Answer in units of A.
Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

Current through the surface at t = 1.1 s is 21.37 A.

Explanation:

The charge is passing through a surface of area varies with time as :

q=q_1t^3+q_2t+q_3

Here,

q_1=5.2\ C/s^3\\\\q_2=2.5\ C/s\\\\q_3=6.5\ C

t is in seconds

q=5.2t^3+2.5t+6.5

The rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(5.2t^3+2.5t+6.5)}{dt}\\\\I=15.6t^2+2.5

At t = 1.1 s, Current,

I=15.6(1.1)^2+2.5\\\\I=21.37\ A

So, the instantaneous current through the surface at t = 1.1 s is 21.37 A.

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Time taken by the arrow to travel along to hit the ground is 0.55 seconds.

Explanation:

The only "force" acting on the "crossbow" to cause it to "hit" the ground is "gravity". There is no initial velocity downward when it shoot.

d=v_{i} t+\frac{1}{2} t^{2}

d = the displacement of the object  

t = the time for which the object moved  

a = acceleration of the object  

v_i = the initial velocity of the object

Given values

d = 1.5 m

t = unknown

a=g=9.8 \mathrm{m} / \mathrm{s}^{2}

\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}

1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}

1.5=0+4.9 \mathrm{t}^{2}

\mathrm{t}^{2}=\frac{1.5}{4.9}

t^{2}=0.306 \mathrm{s}

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How much kinetic energy does an object have that is moving at a rate of 30 m/s and has a mass of 4000 kg ?
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Explanation:

Given that,

The speed of the object, v = 30 m/s

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K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ

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A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after
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Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

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