4 points Two cylindrical resistors R, and R2 made from the same material are connected in parallel in a circuit. Both resistors are at the same temperature when the circuit is operational. Resistor R, has length and diameter d. Resistor R has length 8l and has a diameter d/1. Determine by what factor is the resistance R greater than R.St. your answer as an integer with no decimal place. Type your answer...

Answers

Answer 1

Answer: To determine the factor by which the resistance of resistor R2 is greater than resistor R1, we can use the formula for resistance of a cylindrical resistor:

R = ρ * (L/A)

Where R is the resistance, ρ is the resistivity of the material, L is the length of the resistor, and A is the cross-sectional area of the resistor.

Since both resistors are made from the same material and are at the same temperature, the resistivity ρ is the same for both resistors.

Let's calculate the resistance for R1 and R2 separately.

For resistor R1 (length = L and diameter = d), the cross-sectional area A1 is given by:

A1 = π * (d/2)^2

For resistor R2 (length = 8L and diameter = d/1), the cross-sectional area A2 is given by:

A2 = π * ((d/1)/2)^2

Simplifying the equations:

A1 = π * (d/2)^2 = π * (d^2/4)

A2 = π * ((d/1)/2)^2 = π * (d^2/4)

As we can see, both resistors have the same cross-sectional area A.

Now, let's calculate the resistance for R1 and R2:

R1 = ρ * (L/A1) = ρ * (L / (π * (d^2/4)))

R2 = ρ * (8L/A2) = ρ * (8L / (π * (d^2/4)))

Simplifying the equations further:

R1 = (4ρL) / (πd^2)

R2 = (32ρL) / (πd^2)

Now, let's find the factor by which the resistance R2 is greater than R1:

Factor = R2 / R1 = ((32ρL) / (πd^2)) / ((4ρL) / (πd^2))

Canceling out common terms:

Factor = (32ρL * πd^2) / (4ρL * πd^2)

Factor = 32 / 4

Factor = 8

Therefore, the resistance R2 is 8 times greater than the resistance R1.

Answer 2

Answer:

The resistance of second resistor R2, is 16 times greater than the resistance of the first resistor R1.

The resistance of a cylindrical resistor is given by R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area (which is πd²/4 for a cylinder). For R1, it has length L and diameter d. For R2, it has length 8L and diameter d/1. The resistance of R2 is therefore:

R2 = ρ(8L)/(π(d/1)²/4)

By comparing R2 to R1, we find that R2 is 16 times the resistance of R1.

The resistance of second resistor R2, is 16 times greater than the resistance of the first resistor R1.

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a crowbar having the lenght of 1.75 is used to balance a load of 500N if the distance between the fulcrum and the load is 0.5m find the effort applied to balance the load

Answers

The effort applied = 200 N

Further explanation

Equilibrium :

$\tt F_1.d_1=F_2.d_2$

Total length = 1.75

The distance between the fulcrum and the load is 0.5 m ⇒ d₁=0.5 m

The distance between the fulcrum and the force applied :

$\tt 1.75-0.5=1.25~m\rightarrow d_2=1.25~m$

A load of 500N ⇒ F₁=500 N

The force applied :

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What is the gravitational force acting on a 70.0 kg object standing on the earth's surface ?

Answers

g( gravitational acceleration)=-9.81 m/s^2
m(mass) = 70.0 kg

According to Newton's second law of motion:

"F=ma"

Force is equal mass per acceleration therefore

W=mg
This is because weigth is the force of gravity acting in a body and "a" is replaced by "g" because "g"
is gravitational acceleration

Therefore:

W= mg
W= (70.0 kg) (-9.81m/s^2)
W= -686.7 N

* The "N" means newton which is (kg* m/s^2)

Final answer:

The gravitational force acting on a 70.0 kg object on Earth is calculated using the formula w=mg and is equal to 686 N.

Explanation:

The gravitational force acting on a 70.0 kg object standing on the earth's surface can be calculated using the formula for weight (which is essentially the gravitational force on the object): w = mg, where m is the mass of the object and g is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s².

Therefore, the gravitational force (weight) acting on this object can be calculated as follows: w = mg = (70.0 kg)(9.8 m/s²) = 686 N. Thus, the gravitational force on this 70.0 kg object standing on the earth's surface is 686 Newtons. It's important to understand that this force will vary if the object is moved to a location where g is different, like on the moon.

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True or false. The light bulbs in an office building give off tremendous amounts of heat. Since heat is a form of energy, this heat could be trapped and used to offset our need for electricity, thus saving energy and money.

Answers

It is true that the light bulbs in an office building give off tremendous amounts of heat. This heat can be trapped and used to offset our need for electricity.

Answer:

FALSE

Explanation:

The light bulbs in an office building give off tremendous amounts of heat. Heat is a form of energy, but this heat could not be trapped as heat is a dynamic energy which can not be stored but this energy can flow from high temperature to low temperature.

So here when heat is generated then it will flow towards the region of low temperature and by the time it will convert into several other forms.

So here above statement is NOT correct.

FALSE

Astronauts traveling toward Earth in a fastmovingspacecraft receive a radio signal from an
antenna on Earth. Compared to the frequency
and wavelength of the radio signal emitted from
the antenna, the radio signal received by the
astronauts has a
(1) lower frequency and a shorter wavelength
(2) lower frequency and a longer wavelength
(3) higher frequency and a shorter wavelength
(4) higher frequency and a longer wavelength

Answers

Compared to the frequency and wavelength of the radio signal emitted from
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Answer:

3) Higher frequency and a shorter wavelength

Explanation:

The Doppler Effect occurs when the distance between the wave source and the observer is changing. As the astronauts move toward the source of the radio waves, the radio waves will reach them at a faster rate. The perceived frequency will be higher and the wavelength shorter.

Which has greater buoyant force, light corn syrup or the vegetable oil?

Answers

It is most likely vegetable oil

A driver in a Volkswagen bug (mass = 1335 kg) pulls out going 10 miles per hour in front of a Ford pickup truck (mass = 2675 kg) which is traveling 40 miles per hour. The two vehicles experience a head-on collision. Which of the following statements is true concerning the forces that each vehicle (and driver) experience in this collision? a. The Volkswagen bug experiences a much greater force than the Ford pickup truck due to the greater speed of the truck.
b. The Volkswagen bug experiences a much greater force than the Ford pickup truck due to the larger mass of the truck.
c. The Volkswagen bug experiences a force that is the same as the Ford pickup truck experiences.
d. The Volkswagen bug experiences a much smaller force than the Ford pickup truck due to the smaller speed of the Volkswagen bug.
e. The Volkswagen bug experiences a much smaller force than the Ford pickup truck due to the smaller mass of the Volkswagen bug

Answers

Answer:C

Explanation:

Given

mass of Volkswagen $m_1=1335 kg$

speed of Volkswagen $v_1=10 mi/hr$

mass of Ford pick up $m_2=2675 kg$

speed of ford pick up $v_2=40 mi/hr$

When two bodies collide head-on then force experienced by both the bodies will be the same but opposite in direction.

According to newton's third law, the force exerted by one particle on another is equal to the force exerted by the second particle on first.

Thus two bodies will experience the same force

Answer:

c. The Volkswagen bug experiences a force that is the same as the Ford pickup truck experiences.

Explanation:

Given:

mass of Volkswagen bug, $m_v=1335\ kg$

speed of Volkswagen bug, $v_v=10\ mi.hr^(-1)$

mass of Ford pickup, $m_f=2675\ kg$

speed of Ford pickup, $v_f=40\ mi.hr^(-1)$

Since both undergo a head-on collision they both face the equal impact in accordance with the Newton's third law of motion which states that every action has equal and opposite reaction.

So the bug the force applied by the bug on the pickup is equal to the force applied by the pickup on the bug. This happens only till both the masses are in contact with each other.

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