PHYSICS COLLEGE

Is it true or false if a force is a push or a pull on a object

Answers

Answer 1

Answer:

Answer: true

Explanation: a force can be anything that effects an object, as long as the object moves

Answer 2

Answer: Answer:True

Explanation:Because anything that is being pushed of pulled is called force.

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Enter your answer in the provided box. In water conservation, chemists spread a thin film of certain inert materials over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 32.0 m2 in area. Assuming that oil forms a monolayer (that is, a layer that is only one molecule thick) estimate the length of each oil molecule in nanometers. Assume that oil molecules are roughly cubic. (1 nm = 1 × 10−9 m)

Answers

Answer:

≅3.2 nm

Explanation:

Using the converter units as know for this case that:

1 ml is 1 cubic centimeter ⇒ 0.1 ml is 0.1 cubic centimeters

32.0 m² so :

32.0 m² *100 *100 cm² ⇒ 0.1 / ( 32.0 * 100 *100 ) = 100,000,000 * 0.1 / (32.0 * 100 * 100 ) nm

v = 100/32.0 nm = 3.125 nm thick.

v ≅3.2 nm

As oil is one molecule thick and the molecules are cubic, length of each oil is 3.2 nm

1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building? 2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answers

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1. $s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m$

2. $v = gt = 9.81*3.3 = 32.373 m/s$

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, $v_0$ is the initial velocity, $\Delta s = 48.415$ is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

$0 - v_0^2 = 2*(-9.81)*48.415$

$v_0^2 = 950$

$v_0 = √(950) = 30.82 m/s$

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Starting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up

Answers

Final answer:

The angular acceleration of the disk drive in an old computer game system while speeding up is 1256 rad/s². This is calculated using kinematics in rotational motion, given the information on rotations, revolution time, and start from rest.

Explanation:

To calculate the angular acceleration of a disk drive in an old computer game system, we must use the concept of kinematics in rotation. When it is stated that it takes two revolutions to reach full speed, this implies that the total angular displacement is 4π radians (since one full revolution is 2π radians).

Given that the disk drive revolves once every 0.050 seconds, the final angular speed (ω) can be computed as 2π rad/0.050 s = 125.6 rad/s. Since the disk starts from rest, the initial angular speed (ω0) is 0. As a result, the total time taken (t) to reach full speed is 2*0.050s = 0.1 s.

We can then use the equation of motion in rotational form, α = (ω - ω0)/t, to calculate the angular acceleration. Hence the angular acceleration (α) is (125.6 rad/s - 0 rad/s) / 0.1 s = 1256 rad/s². Therefore, the angular acceleration of the disk drive is 1256 rad/s² while it is speeding up.

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Final answer:

The angular acceleration of the disk drive while it is speeding up is 8π rad/s².

Explanation:

The angular acceleration of the disk drive while it is speeding up can be determined by using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time taken. In this case, the initial angular velocity is 0 (since the disk starts from rest) and the final angular velocity is 2 revolutions per 0.050 seconds. To convert revolutions to radians, multiply by 2π. The time taken is the time for two revolutions, so it is 2 * 0.050 seconds. Plugging in these values in the formula, we get:

Angular acceleration = (2 * 2π rad/s - 0) / (2 * 0.050 s) = 8π rad/s²

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Two loudspeakers, 4.0 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m.a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answers

Answer:

a) 343.0 Hz b) 686.0 Hz

Explanation:

a) First, we need to know the distance to both speakers.

If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.

So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:

d₁ = 2.0 m-0.25 m= 1.75 m

d₂ = 2.0 m + 0.25 m = 2.25 m

The difference between these distances is the path difference between the sound from both speakers:

d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m

If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:

d = (2*n-1)*(λ/2) = 0.5 m

The minimum distance is for n=1:

⇒ λ = 2* 0.5 m = 1 m

In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:

v = λ*f, where v= 343 m/s and λ=1 m.

Solving for f, we have:

$f =(343.0 m/s)/(1.0 m) = 343 Hz$

b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:

d = 0.5 m = 2n*(λ/2) = λ (for n=1)

if the speed remains the same (343 m/s) we can find the new frequency as follows:

$f =(v)/(d) =(343 m/s)/(0.5m) =686.0 Hz$

⇒ f = 686.0 Hz

Final answer:

Two speakers create peaks and troughs of sound intensity due to constructive and destructive interference of waves. Using wave properties, the frequency of the sound when a minimum intensity is experienced 0.25m from the center is 680Hz. Increasing the frequency, the first to produce maximum intensity at the same position is about 2720Hz.

Explanation:

The behavior of sound intensity in this question is due to wave interference, specifically, constructive and destructive interference of sound waves. When you stand halfway between the speakers, the sound waves from each speaker are in phase, which means the pressure variations combine to create an intensified sound, known as constructive interference.

When you move towards one of the speakers and detect a minimum in sound intensity, this is due to destructive interference, which occurs when the crest of one wave overlaps with the trough of another, canceling each other and producing a minimum sound level.

a. The frequency of the sound can be calculated using the formula for wave speed, v = f.lambda, where v is the speed of sound (340 m/s under normal conditions), f is the frequency, and lambda is the wavelength. In this case, a minimum sound intensity indicates one-half wavelength. So, lambda = 0.5 m. Thus, frequency, f = v/lambda = 340/0.5 ~ 680 Hz.

b. When you increase the frequency while remaining 0.25m from the center, the first frequency for which the location will be a maximum of sound intensity will be when you are an integral multiple of the wavelength away from the source. Thus if we let this be 2λ, we can calculate the frequency as f = v / λ = v / (0.25m / 2) = 340 / 0.125 ~ 2720 Hz.

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Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?

Answers

Answer:

The electric force increases by a factor of 4.

Explanation:

The electric force between two charges $q_1$ and $q_2$ separated a distance d can be calculated using Coulomb's Law:

$F=(kq_1q_2)/(d^2)$

where $k=9*10^9Nm^2/C^2$ is the Coulomb constant.

If the value of each charge is doubled, then we will have a force between them which is:

$F'=(k(2q_1)(2q_2))/(d^2)=4(kq_1q_2)/(d^2)=4F$

So the new force is 4 times larger than the original force.

Final answer:

Doubling the charge on each particle increases the electric force between them by a factor of 4.

Explanation:

The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.

Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.

So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.

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A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?