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Fantom [35]
2 years ago
6

A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw b

ased on this description?
A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
B. The electric field points to the right because the force on a negative charge is in the same direction as the field.
C. No conclusion can be drawn because the sign of the charge creating the field is unknown.
D. No conclusion can be drawn because the amount of charge on the test charge is unknown.
Physics
2 answers:
lapo4ka [179]2 years ago
4 0
The best conclusion to draw based on the description would be: <span>A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
This phenomenon happened because </span><span>The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge</span>
Marrrta [24]2 years ago
4 0

Answer:

A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

As we know that electrostatic force on a charge in external electric field is given as

\vec F = q\vec E

so here the force and electric field will be in same direction if charge is positive and if the charge is negative then force will be in opposite direction to the electric field.

So we know that since the electron is negatively charged so the force on electron must be opposite to the electric field present in that region.

So correct answer will be

A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

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conserved

Explanation:

During this process the energy is conserved

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a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll
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Answer:

h = 0.0362\,m

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

U_{g,A} + K_{A} = U_{g,B} + K_{B}

U_{g,B} - U_{g,A} = K_{A} - K_{B}

m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}

The moment of inertia of the coin is:

I = \frac{1}{2}\cdot m \cdot r^{2}

After some algebraic handling, an expression for the maximum vertical height is derived:

m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}

h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}

h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

h = 0.0362\,m

3 0
2 years ago
Water is kept in a vessel at a temperature of 100°C. What would happen if a metal ball having a temperature of 30°C is dropped i
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Answer:

there will be a heat flow from water to the metal ball...

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2 years ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
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To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of
Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = 20\times 2000\ lb

d = Diameter = 1.25 in

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L_2 = Final length

A = Area = \dfrac{\pi}{4}d^2

Y = Young's modulus = 30\times 10^6\ psi

Young's modulus is given by

Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft

The length during the lift is 50.0543248872 ft

6 0
2 years ago
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