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2 years ago

What happens if : . The test charge is not tiny.

1 answer:

3 0

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

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Despite the gases, dust, and stars that compose the universe, it is still mostly made up of ____________________.

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Answer:

space

Explanation:

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3 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

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3 years ago

Equation of motion description is v=20+2t.How big is the initial speed,acceleration?

trasher [3.6K]

Answer:

the initial velocity is 20 m/s and the acceleration is 2 m/s²

Explanation:

Given equation of motion, v = 20 + 2t

If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;

From first kinematic equation;

v = u + at

where;

v is the final velocity

u is the initial velocity

a is the acceleration

t is time of motion

If we compare (v = u + at) to (v = 20 + 2t)

then, u = 20 and

a = 2

Therefore, the initial velocity is 20 m/s and the acceleration is 2 m/s²

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3 years ago

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

<u>Now the power can be given by:</u>

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

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3 years ago

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080

deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

$I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}$

now we have

$M = 0.120 kg$

$m_1 = 0.02 kg$

$m_3 = 0.08 kg$

now we have

$I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}$

so we have

$I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162$

$I = 0.02835$

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

$\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}$

$\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)$

$\omega = 4.32 rad/s$

Now speed of 0.08 kg mass when it reaches to bottom point is given as

$v = \omega \frac{L}{2}$

$v = 4.32 (0.45)$

$v = 1.94 m/s$

3 0

3 years ago