Select all the statements regarding electric field line drawings that are correct. Group of answer choices:
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.
Answers
Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.
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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.
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If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity level. A. 1.6 WB. 0.06 WC. 97 dBD. 223 dBE. 179 dB
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.b. What should be its sign, so that all three charges will be in equilibrium?c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
Answers
Answer:
His results gave the first evidence that atoms were made up of smaller particles.
Answers
Answer:
Explanation:
From the question we are given;
Acceleration a = 6.07m/s²
Time t= 0.25s
Final velocity v = 9.64m/s
Required
Initial velocity u
Using the equation of motion
v = u+at
9.64 = u+(6.07)(0.25)
9.64 = u+1.5175
u = 9.64-1.5175
u = 8.1225m/s
Hence the object's initial velocity is 8.1225m/s
48 degrees
57 degrees
61 degrees
Answers
Answer:
61 degrees, I just did the test.
Explanation:
Answer: 61 degrees
Explanation:
I just did the question and got it right
0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts
Answers
The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.
What is electric power?
Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.
The given data in the problem is;
V is the voltage = ? Volt (V)
Electric current (I)= 15.8 amps (A)
P is the power =10.0 kilowatts =10⁴ watt
The formula for the power is given as;
The voltage of the electricity will be 63.29 V.
Hence, option B is correct.
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Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.
10,000 W / 15.8 A = V
632.9, or 633.
Answers
Explanation:
Below is an attachment containing the solution.
Answers
Answer:
The correct answer is "21195 N".
Explanation:
The given values are:
Tensile strength,
= 3000 MN/m²
Diameter,
= 3.0 mm
i.e.,
= 3×10⁻³ m
Now,
The maximum load will be:
=
On substituting the values, we get
=
=
=
Final answer:
The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.
Explanation:
The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.
We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.
Learn more about Tensile Strength here:
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