Ask question

Ask question

All categories

1 month ago

8

Ellen does an experiment by releasing a ball from a height of 1 m above each floor in a tall building. She records the time it t

akes the ball to reach the floor.
Which question is Ellen most likely investigating?

How does distance affect the gravitational force on objects?
How does distance affect the electromagnetic force on objects?
Does height affect the strong force acting on a moving ball?
Does height affect the weak force acting on a moving ball?

2 answers:

serg [3.5K]1 month ago

7 0

The question Ellen is likely exploring is "In what way does distance influence the gravitational force acting on objects?"

Explanation:

inna [3.1K]1 month ago

4 0

The correct answer is A since I answered the question accurately

You might be interested in

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

By dividing equation (4) by (3),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

According to equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

This results in a 3.5 fold increase in energy.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy when the plate separation transitions to d₂ can be written as:

$E_2=\frac{1}{2} C_2V^2$ .....(6)

Referencing equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this particular scenario,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

6 0

1 month ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

kicyunya [3294]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak $mg-T=\frac{mv^2}{r}$

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as $mg-0=\frac{mv^2}{r}$

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at the bottom $E_b=\frac{mv_0^2}{2}$

By comparing the two states using conservation of energy, we find $v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

6 0

2 months ago

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

kicyunya [3294]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For the maximum emf, $\sin\omega t=1$

Therefore,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

Hence, the maximum emf generated in the loop is 5.32 V.

3 0

2 months ago

In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c

Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

$GDP=\pi } \frac{d^{2} }{4}$

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

$GDP_{b} =\pi \frac{3^{2} }{4}$

Conversely, in darkness, the diameter typically enlarges to 7 mm.

$GDP_{b} =\pi \frac{7^{2} }{4}$

This indicates an increase in light-gathering capacity.

$Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}$

Thus, the amount of light the eye can capture is 31.4 mm².

3 0

2 months ago

These little bunnies (A, B, and C) were born in the same litter to the same parents, but they have different traits for the feat

Ostrovityanka [3204]

Response/Clarification:

Each of us receives 2 versions of a gene from our parents, with one inherited from the mother and one from the father.

Both our mother and father possess 2 versions of every gene. Therefore, the specific version we inherit is determined randomly, much like the outcome of a coin flip. This applies to both parents.

For instance, if the mother has one variant causing thick ears (A) and another for thin ears (a), she is Aa.

Similarly, the father also has these variants and is Aa as well.

The father can transfer either A or a, while the mother can also transfer either A or a.

As a result, their offspring can be AA, Aa, or aa. An AA genotype results in thick ears, while aa results in thin ears. The Aa genotype produces ears of intermediate thickness, akin to bunny B. This demonstrates the concept of incomplete dominance

8 0

1 month ago