PHYSICS HIGH SCHOOL

The current in the circuit shown is 2.0 A.What is the value of R3?

A) 10 Ω
B) 15 Ω
C) 20 Ω
D) 55 Ω

Answers

Answer 1

Answer:

Answer:

Resistor R₃ = 10 Ω

Explanation:

It is given that,

Current flowing in the circuit, I = 2 A

Resistors R₁ = 2 Ω

Resistor R₂ = 3 Ω

Voltage, V = 30 V

We have to find the value of R₃. All three resistors are connected in series. In series combination, the current flowing through the all resistors is same. Firstly, calculating equivalent resistance of three resistors as :

$R_(eq)=R_1+R_2+R_3$

$R_(eq)=2\ \Omega+3\ \Omega+R_3$

$R_(eq)=5\ \Omega+R_3$ ..............(1)

Using Ohm's law : $V=IR_(eq)$

$30\ V=2\ A* (5 + R_(3))$

On solving above equation : $R_3=10\ \Omega$

Hence, the correct option is (A) " 10 Ω "

Answer 2

Answer:

The value of R3 is A) 10 Ω

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The process is "transmission" of light. The object is "transparent".

When light passes directly through an object unchanged, that object is called transparent.
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Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete.
(b) On wet concrete.
(c) On ice, assuming that μs = 0.100 , the same as for shoes on ice.

Answers

Final answer:

The maximum acceleration of a car moving uphill can be calculated using the formula μs*g*cosθ - g*sinθ where θ is the slope angle, μs is the coefficient of static friction, and g is the acceleration due to gravity. The figures for μs differ depending on the road condition - dry concrete, wet concrete, or ice, substantially affecting the car's acceleration.

Explanation:

The maximum acceleration of a car moving uphill is determined by the force of static friction, which opposes the combined force of the car's weight component down the plane and the force utilized by the driving wheels. The maximum static friction force (F_max) is determined by the coefficient of static friction (μs) multiplied by the normal force (N), which is equivalent to the weight of the car (mg) multiplied by the cosine of the angle (cosθ).

(a) On dry concrete: Since the μs is usually 1.0 on dry concrete and half the weight of the car is supported by the drive wheels, the maximum acceleration can be calculated as μs*g*cosθ - g*sinθ

(b) On wet concrete: The μs is around 0.7 on wet concrete. Substituting this value into the formula would give us the maximum acceleration on a wet surface.

(c) On ice: With a μs value of 0.1 as given, the maximum acceleration on ice can also be calculated using the same formula.

As one can see, the road conditions significantly impact the car's maximum acceleration due to the change in the amount of friction between the tires and road surface.

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Final answer:

The maximum accelerations for the car going up a 4º slope are 9.3 m/s² on dry concrete, 6.4 m/s² on wet concrete, and -0.1 m/s² on ice.

Explanation:

The maximum acceleration of the car up the slope can be calculated using the equation: a = μs * g * cosθ - g * sinθ, where a is the acceleration, μs is the coefficient of static friction, g is the acceleration due to gravity, and θ is the angle with the horizontal.

To solve this problem, we must teach the student to take several factors into account, including the various coefficients of static friction corresponding to different road conditions, namely dry concrete, wet concrete, and ice.

Considering that each scenario has different values of μs, we fill in the equation with the angles and coefficients of static friction. As a rule of thumb, μs for dry concrete is generally taken as 1.0, for wet concrete as 0.7 and for ice (mentioned in the question) as 0.100.

For dry concrete, a = 1.0 * 9.8 * cos(4) - 9.8 * sin(4) = 9.3 m/s².
For wet concrete, a = 0.7 * 9.8 * cos(4) - 9.8 * sin(4) = 6.4 m/s².
For ice, a = 0.100 * 9.8 * cos(4) - 9.8 * sin(4) = -0.1 m/s².

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Describe the difference between the currents that exist in the wires leading to a capacitor when these wires are connected to (a) a DC source and (b) an AC source.

Answers

Answer:

Explanation: In DC circuit, the current will flow for a short time, which is required to charge the capacitor. Once you switch it on, it spikes and the gradually decreases to almost zero (0) as the capacitor becomes fully charged.

In an AC circuit, the circuit acts as if the current is flowing throw the plates whereas is not actually flowing. The circuit acts like the AC is flowing through the capacitor.

At a given location in space, the magnetic field in an electromagnetic wave is increasing. How is the electric field changing at that same location? a. The electric field is increasing.
b. The electric field is decreasing.
c. The electric field is not changing.

Answers

Answer:

Option a) is correct

Explanation:

In the given question option a) ( i.e the electric field is increasing ) is correct.

The above option is correct because the electric field and the electromagnetic wave both are in the same phase of the EM wave. Thus, the electromagnetic wave and the electric field are directly proportional to each other.

Final answer:

In an electromagnetic wave, the electric and magnetic fields are linked, oscillating in phase. Therefore, when the magnetic field is increasing, the electric field is also increasing.

Explanation:

In an electromagnetic wave, the electric field and the magnetic field are perpendicular to each other and they oscillate in phase. This means when the magnetic field is increasing, the electric field is also increasing, and vice versa. So, the correct answer is option a. The electric field is increasing. It's important to keep this correlation in mind while studying electromagnetic waves, as these oscillations and their interplay are fundamental to how these waves propagate through space.

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Where does sugar enter the blood?

Answers

After we eat, blood sugar concentratons rise, the pancreas releases insulin automatically so that the glucose enters cells, as more and more cells receive glucose, blood sugar levels come down to normal again. Exces glucose is stored as glycogen in the liver and muscles.Hope this helps.

A 17,000-kg airplane lands with a speed of 82 m/s on a stationary aircraft carrier deck that is 115 m long. find the work done by nonconservative forces in stopping the plane

Answers

Final answer:

The work done by nonconservative forces in stopping the 17,000-kilogram airplane landing at a speed of 82 m/s is 57,062,000 Joules. This is calculated by the change in kinetic energy of the airplane when it lands and comes to a stop.

Explanation:

The question refers to the concept of work-energy theorem in Physics, especially involving non-conservative forces. The airplane is initially moving and finally comes to rest. Its initial kinetic energy (KE) gets transferred to work done by nonconservative forces, which in this scenario includes friction due to the aircraft carrier deck and air resistance.

The initial kinetic energy of the plane is calculated using the formula 1/2 * m * v^2 where 'm' is the mass of the plane and 'v' is its speed. So, the initial kinetic energy of the plane is 1/2 * 17,000 kg * (82 m/s)^2 = 57,062,000 Joules. When the plane comes to rest, its final kinetic energy is 0. As per the work-energy theorem, the work done by nonconservative forces is equal to the change in the kinetic energy. Therefore, the work done by nonconservative forces in stopping the plane = Initial KE - Final KE = 57,062,000 Joules - 0 = 57,062,000 Joules.