(True or False) A small force exerted over a large time interval can create the same change in momentum as a large force exerted over a small time interval. *

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Answer 1

A small force exerted over a large time interval can indeed create the same change in momentum as a large force exerted over a small time interval. The statement is True.

The concept that relates force, time, and momentum is known as impulse. Impulse is the product of force and time, and it is equal to the change in momentum experienced by an object.

Impulse = Force × Time

By rearranging this equation, we can see that for a given change in momentum, if the force acting on an object is smaller, the time over which the force is applied will be longer, and vice versa. This demonstrates the principle of conservation of momentum.

As long as the product of force and time remains the same, the change in momentum will be equivalent.

Therefore, a small force exerted over a large time interval can indeed produce the same change in momentum as a large force exerted over a small time interval.

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Related Questions

The 17th century astronomer who kept a roughly 20 year continuous record of the positions of the Sun, Moon, and planets was: Group of answer choices

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The 17th-century astronomer who kept a roughly 20-year continuous record of the positions of the Sun, Moon, and planets was Johannes Hevelius.

Hevelius was a Polish astronomer, mathematician, and brewer who made significant contributions to the field of astronomy during the 17th century. He meticulously observed and recorded the positions of celestial objects, publishing his observations in his monumental work titled "Prodromus Astronomiae" in 1690. This work contained a detailed star catalog, lunar maps, and records of planetary positions, including those of the Sun and Moon.

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nearsightedness and farsightedness can be corrected with the use of: eyeglasses contact lenses vitamin a eye drops

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Eyeglasses and contact lenses are the primary methods used to correct nearsightedness and farsightedness. While vitamin A is important for overall eye health, it does not directly correct these vision problems. Eye drops are not used for correcting these refractive errors.

Nearsightedness and farsightedness are two common vision problems that can be corrected with the use of different methods. Let's discuss each correction option:

1. Eyeglasses: Eyeglasses are the most common and effective method for correcting both nearsightedness and farsightedness. In the case of nearsightedness, the lenses of the glasses are concave, which helps to diverge the incoming light rays before they reach the eye, allowing the image to be focused properly on the retina. For farsightedness, the lenses are convex, which converges the light rays and helps to focus the image on the retina. Eyeglasses provide a simple and non-invasive solution, and they can be easily adjusted to suit an individual's prescription.

2. Contact lenses: Contact lenses also provide an effective correction option for both nearsightedness and farsightedness. These are small, thin lenses that are placed directly on the surface of the eye. They work in a similar way to eyeglasses by altering the path of light entering the eye. Contact lenses offer a wider field of view compared to glasses and are generally more suitable for individuals who are involved in sports or other physical activities.

3. Vitamin A: While vitamin A is important for overall eye health, it does not directly correct nearsightedness or farsightedness. However, a deficiency in vitamin A can contribute to certain eye conditions, such as night blindness. Therefore, maintaining a healthy diet that includes foods rich in vitamin A, such as carrots and leafy greens, is important for good eye health.

4. Eye drops: Eye drops are typically used for treating dry eyes or eye infections and are not directly related to correcting nearsightedness or farsightedness.


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Suppose a laser beam is projected downward through the air and is incident upon a face of a right triangular prism that has an index of refraction of 2.75. Find (A) the refracted angle of the light (B) whether the beam will hit the bottom surface or the right-hand surface (C) What will happen when the light hits the surface you indicated in (B) -- will it be internally reflected or refracted into the air? Show this with calculations.

Answers

A) To find the refracted angle of the light, we can use Snell's law which states that n1*sin(theta1) = n2*sin(theta2), where n1 and n2 are the indices of refraction of the two mediums, and theta1 and theta2 are the angles of incidence and refraction respectively.

In this case, the air has an index of refraction of 1, and the prism has an index of refraction of 2.75. Let's assume the angle of incidence is theta1.
Using Snell's law, we have: 1*sin(theta1) = 2.75*sin(theta2)
Rearranging the equation, we get: sin(theta2) = (1/2.75)*sin(theta1)
To find theta2, we take the inverse sine of both sides: theta2 = sin^(-1)((1/2.75)*sin(theta1))
B) To determine whether the beam will hit the bottom surface or the right-hand surface, we need to consider the critical angle. The critical angle is the angle of incidence at which the refracted angle becomes 90 degrees.
Using Snell's law, we have: 1*sin(critical angle) = 2.75*sin(90)
Simplifying, we find: sin(critical angle) = 2.75
Taking the inverse sine, we get: critical angle = sin^(-1)(2.75)
If the angle of incidence is greater than the critical angle, the light will be totally internally reflected and hit the right-hand surface. Otherwise, it will hit the bottom surface.
C) When the light hits the surface indicated in (B), if the angle of incidence is greater than the critical angle, it will be totally internally reflected. If the angle of incidence is less than the critical angle, it will be refracted into the air.
Please note that to provide specific calculations, the values of theta1 and the critical angle are needed.

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Determine the orbital period for an object orbiting at a distance of 7.3x10^8 from the center of a spherical object whose mass is 3.0x10^27 at a velocity of 2.8x10^4.

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The orbital period for an object can be determined using Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the average distance from the center of the spherical object.

To calculate the orbital period, we can use the formula:

[tex]T^2 = (4π^2 / G * M) * r^3[/tex]
Where T is the orbital period, G is the gravitational constant[tex](6.67430 × 10^-11 m^3 kg^-1 s^-2)[/tex], M is the mass of the spherical object, and r is the distance from the center of the spherical object.

Given:
Distance from the center of the spherical object, r = 7.3x[tex]10^8[/tex] m
Mass of the spherical object, M =[tex]3.0x10^27[/tex] kg

First, we need to calculate [tex]T^2[/tex]using the given values:

[tex]T^2 = (4π^2 / G * M) * r^3[/tex]

Plugging in the values:
[tex]T^2 = (4 * π^2 / (6.67430 × 10^-11 m^3 kg^-1 s^-2) * (3.0x10^27 kg)) * (7.3x10^8 m)^3[/tex]
Simplifying the equation:
[tex]T^2 = (4 * π^2 / (6.67430 × 10^-11 m^3 kg^-1 s^-2)) * (3.0x10^27 kg) * (7.3x10^8 m)^3[/tex]

Calculating [tex]T^2:[/tex]
[tex]T^2 = 1.75x10^20 s^2 * (3.0x10^27 kg) * (7.3x10^8 m)^3[/tex]
[tex]T^2 = 2.39x10^62 m^3 kg^-1 s^-2[/tex]

Now, we can find the orbital period T by taking the square root of[tex]T^2[/tex]:

[tex]T = sqrt(2.39x10^62 m^3 kg^-1 s^-2)[/tex]

Therefore, the orbital period for the object is approximately sqrt(2.39x10^62) seconds.

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You are checking the calibration of a treadmill at 3.5mph. when you calculate the speed,you calculate 3.5 mph. this indicates the treadmill is:_________

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You are checking the calibration of a treadmill at 3.5mph. when you calculate the speed, you calculate 3.5 mph. this indicates the treadmill is accurate.

The correct term to fill in the blank is "accurate." When you calculate the speed of the treadmill and obtain a measurement of 3.5 mph, it indicates that the treadmill is calibrated correctly and providing an accurate speed reading. Calibrating a treadmill involves ensuring that it accurately measures the speed at which it is moving. In this case, the treadmill's measurement aligns with the intended speed of 3.5 mph, confirming that it is properly calibrated.

By verifying the accuracy of test equipment, calibration aims to minimize any measurement uncertainty. In measuring procedures, calibration quantifies and reduces mistakes or uncertainties to a manageable level.

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you’re in tucson and you notice a star that’s rising in the southeast (azimuth >90). how long will it be before this star sets?

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If the star is currently rising in the southeast (azimuth > 90 degrees), it will take approximately 6 hours for it to set

The time it takes for a star to set after it has risen in the southeast depends on several factors, including the star's declination, the observer's latitude, and the current time of the year. In Tucson, which is located at a latitude of approximately 32 degrees North, stars with a declination greater than 58 degrees will never set below the horizon.

Assuming the star has a declination that allows it to set, we can estimate the time it takes for it to set by considering the rotation of the Earth. On average, the Earth rotates 15 degrees per hour, which corresponds to one hour for every 15 degrees of azimuth.

If the star is currently rising in the southeast (azimuth > 90 degrees), it will take approximately 6 hours for it to set in the southwest (azimuth = 180 degrees) if we assume a constant rate of rotation. However, this is a rough estimation and may vary depending on the specific circumstances.

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_________________ was the first astronomer to make telescopic observations which demonstrated that the ancient Greek geocentric model was false.

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Galileo Galilei was the first astronomer to make telescopic observations that demonstrated that the ancient Greek geocentric model was false. He was a renowned Italian astronomer, mathematician, and physicist of the seventeenth century.

He was a key figure in the Scientific Revolution, advocating for a scientific method that emphasized experimentation and observation, which differed from the traditional Aristotelianism that had dominated scientific thinking for centuries.Galileo made important contributions to the fields of astronomy and physics. He invented an improved telescope that enabled him to observe the sky more clearly than any astronomer had before him.

Through his telescope, Galileo observed the phases of Venus, the four largest moons of Jupiter, the rings of Saturn, and sunspots, among other things. These discoveries provided evidence for the heliocentric model of the solar system, which proposed that the Earth and other planets revolve around the sun, rather than the Earth being the center of the universe, as had been previously believed.

Galileo’s ideas and observations were met with significant opposition, particularly from the Catholic Church, which viewed his work as a threat to the church’s traditional teachings. In 1633, Galileo was tried by the Inquisition, found guilty of heresy, and placed under house arrest for the remainder of his life. Despite the persecution he faced, Galileo’s work laid the foundation for the modern scientific method and revolutionized our understanding of the universe.

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A for loop is used when a loop is to be executed a known number of times.

a. true

b. false

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For loop is used when a loop is to be executed a known number of times, it is TRUE.

For loop is indeed used when a loop is to be executed a known number of times. In programming, the for loop is a control structure that allows repeated execution of a block of code based on a specified condition. It consists of three main components: initialization, condition, and increment/decrement. The loop executes as long as the condition is true and terminates when the condition becomes false.

The for loop is particularly useful when the number of iterations is predetermined or known in advance. By specifying the initial value, the loop condition, and the increment/decrement, we can control the number of times the loop body will be executed. This makes it a suitable choice when a specific number of iterations or a well-defined range needs to be handled.

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The molecule that functions as the reducing agent in a redox reaction ___ electrons and ______ energy.

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The molecule that functions as the reducing agent in a redox reaction gains electrons and releases energy.

Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states. The term ‘redox’ is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes: a reduction process and an oxidation process.

The oxidation and reduction reactions always occur simultaneously in redox or oxidation-reduction reactions. The substance getting reduced in a chemical reaction is known as the oxidizing agent, while a substance that is getting oxidized is known as the reducing agent.

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A closely wound, circular coil with a diameter of 3.40 cm has 410 turns and carries a current of 0.600 A .

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1) The magnitude of the magnetic field at the center of the coil is 0.0609 T. 2) The magnitude of the magnetic field at a point on the axis of the coil a distance of 8.20cm from its center is [tex]7.82 * 10^{-6} T[/tex]

1) The magnetic field at the center of the coil can be calculated using the formula:

[tex]B = \mu_0 * (N * I) / (2 * R)[/tex],

where  [tex]\mu_0[/tex] is the permeability of free space [tex](4\pi * 10^{-7} T.m/A)[/tex], N is the number of turns in the coil (410), I is the current flowing through the coil (0.600 A), and R is the radius of the coil (half the diameter, 3.40 cm/2 = 1.70 cm = 0.017 m).

Plugging in these values:

[tex]B = (4\pi * 10^{-7} T.m/A) * (410 * 0.600 A) / (2 * 0.017 m) = 0.0609 T[/tex]

2) For calculating the magnetic field at a point on the axis of the coil, a distance of 8.20 cm from its center, we can use the formula:

[tex]B = \mu_0 * (N * I * R^2) / (2 * (R^2 + d^2)^(3/2))[/tex],

where d is the distance of the point from the center of the coil (8.20 cm = 0.082 m).

Plugging in the values:

[tex]B = (4\pi * 10^{-7} T.m/A) * (410 * 0.600 A * (0.017 m)^2) / (2 * ((0.017 m)^2 + (0.082 m)^2)^(3/2)) = 7.82 * 10^{-6} T[/tex]

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The complete question is:

A closely wound, circular coil with a diameter of 3.40 cm has 410 turns and carries a current of 0.600A

1) What is the magnitude of the magnetic field at the center of the coil?

2) What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 8.20cm from its center?

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