What happend to the force of gravity between two subjects if the distance between them decreases?

300,000 kilo

your welcome

Answer:

Tripled

Explanation:

The relation between the tangential speed and the angular speed is given by

v = r ω

where, r is the radius of the circular path and ω be the angular velocity

As we know that

ω = 2π f

Where, f is the frequency of rotation in rpm

So, v = 2π f r

As the RPM becomes tripled, the tangential velocity is also tripled.

Answer:to the right

Explanation:

A.

The rate of change of the area of a circle with a decreasing radius at the instant the radius is 10ft, is -40π ft²/s.

The subject of this question seems to be about the concept of rate of change specifically dealing with the changing area of a circle given a decreasing radius. The circle's area is defined by the equation A = πr². We are told the radius is decreasing at a rate of 2ft/s. To find the rate of change of the area, we can use the formula for the derivative of the circle's area with respect to time, which is dA/dt = 2πr(dr/dt). Plugging in 10 for the radius (r) and -2 for dr/dt (since the radius is decreasing), we get dA/dt = 2π*10*(-2) = -40π ft²/s. So, at the instant the radius is 10ft, the area is decreasing at a rate of 40π ft²/s.

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The rate of change of the area of a circle at the instant the radius is 10ft, while the radius is decreasing at 2ft/s, is -40π ft²/s or putting π = 3.14 we get -125.6 ft²/s.

The question is asking for the rate of change of the area of a circle when the radius is decreasing at a constant rate. Since the area of a circle is defined by the equation A = πr² , where A is the area and r is the radius, we use calculus to find the rate of change of the area with respect to time ([tex]\frac{dA}{dt}[/tex]).

When the radius r is 10ft, the derivative of the area with respect to the radius ([tex]\frac{dA}{dt}[/tex]) is 2πr.

Since the question states that the radius is decreasing at a rate of 2ft/s, we have :

[tex]\frac{dr}{dt}[/tex] = -2ft/s.

Using the chain rule, we can find the rate of change of the area by multiplying these two rates:

[tex]\frac{dA}{dt}[/tex] = [tex]\frac{dA}{dr}[/tex] x [tex]\frac{dr}{dt}[/tex].

So at r = 10ft,

[tex]\frac{dA}{dt}[/tex] = 2π x 10ft x -2ft/s,

or, [tex]\frac{dA}{dt}[/tex] = -40π ft²/s .

putting π = 3.14 we get

[tex]\frac{dA}{dt}[/tex] = -125.6 ft²/s.

To find acceleration using velocity and distance, use the kinematic equation v² = vo² + 2a(d − do). If the object starts from rest, rearrange the equation to solve for acceleration: a = (v² - vo²)/2d. Keep in mind that this approach assumes constant acceleration and ignores forces such as air resistance and friction.

To find acceleration when given distance and velocity, you can use physics' kinematic equations that relate displacement, velocity, acceleration, and time. One of these equations is v² = vo² + 2a(d − do), where v is the final velocity, vo is the initial velocity, a is the acceleration, and d is the displacement or distance.

First, to find acceleration (a), you have to ensure that the initial velocity (vo) is zero (when the object starts from rest). After this, rearrange the equation to solve for acceleration: a = (v² - vo²)/(2d). This will help you find the acceleration when you plug in your given final velocity (v) and distance (d).

Note: This method assumes a constant acceleration and does not consider forces like friction or air resistance. It's fundamental to consider this when dealing with real-world problems.

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It is "c", and i hope it's correct. please pm me, if you need more help

Color is the frequency of the light and wavelength is the speed of that light. Depending on the color it will have a higher frequency and a smaller wavelength while another will have a lower frequency and a longer wavelength so that the colors will even out. Hope this helps.

The color of light is determined by its wavelength. In the visible light spectrum, longer wavelengths are perceived as red, while shorter wavelengths are perceived as blue or violet. This is evident in phenomena like rainbows.

The relationship between color and wavelength for light is that the color of light that we perceive is determined by the wavelength of the light. In the visible light spectrum, each color corresponds to a different range of wavelengths. For instance, red light has longer wavelengths (about 700 nM to 635 nM), whereas blue or violet light has shorter wavelengths (around 450 nM to 400 nM). This is why when we see a rainbow, it is a gradation of colors from red (long wavelength) at the top to violet (short wavelength) at the bottom. This happens because of the refraction and dispersion of light in droplets of water.

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Planck's equation can be used as a formula to determine the energy of a photon. In this problem, given the frequency and using 6.63 x 10^-34 J.s as Planck's constant, the energy can be calculated. This Physics concept is useful in understanding the energy-frequency relationship of photons.

To calculate the energy of a photon, we can use the Planck's equation E = hv, where 'E' is the energy of the photon, 'h' is the Planck's constant and 'v' is the frequency of the photon.

In this case, the frequency of the photon 'v' is given as  3.55 x 10^17 Hz and Planck's constant 'h' is a universally accepted constant which is 6.63 x 10^-34 J.s. Substituting these values into Planck's equation we get:

E = 6.63 x 10^-34 J.s x 3.55 x10^17 Hz

Upon calculating, the energy in Joules of the photon can be found.

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Answer:

Cancer

Explanation:

First calculate for the time it takes for the model rocket to reach the land. Using the formula:

h = vi t + 0.5 g t^2

where h is height = 100 m, vi is initial vertical velocity = 0 since it simply dropped, t is time = ?, g = 9.8 m/s^2

100 = 0.5 (9.8) t^2

t = 4.52 seconds

So the total horizontal distance taken is:

d = 50 m/s * 4.52 s

d = 225.90 m

The distance from the edge of the cliff where the model rocket will land is ; 225.90 m

Using the given data :

Velocity of rocket = 50.0 m/s

Depth of canyon ( H ) = 100 m deep

Initial velocity ( vi ) = 0

To determine how far from the edge of the cliff does the model rocket land

H = vi( t ) + 0.5gt²

100 = 0.5*(9.8)*t²

∴ t = 4.52  secs

Velocity = horizontal distance / time

∴ distance ( horizontal distance ) = velocity * time

   distance = 50 m/s * 4.52 secs = 225.90 m

Hence we can conclude that The distance from the edge of the cliff where the model rocket will land is ; 225.90 m .

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To find the mass of the crate on a frictionless incline with a known force and acceleration, use Newton's second law. The mass is calculated as 40 N divided by 2.0 m/s^2, resulting in 20 kg.

To determine the mass of the crate on an incline, you can use Newton's second law of motion, which is F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, we know the force applied up the incline (40 N), the acceleration of the crate (2.0 m/s2), and we need to determine the mass m. The equation can be rearranged to solve for mass by dividing both sides by acceleration, resulting in m = F/a. Thus, dividing the force by the acceleration, you get m = 40 N / 2.0 m/s2 = 20 kg. So, the mass of the crate is 20 kilograms.

Answer:

TITE

Explanation:

in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

Answer:

B:when its tangential speed is constant

Although the speed of an object that has a uniform circular motion is constant, its velocity is not constant. Not only that, but it is actually changing constantly.

The formula for calculating acceleration can be derived from:

v^2 = v0^2 + 2 a d

where a is acceleration, v is final velocity, v0 is initial velocity and d is distance

Rewriting in terms of a:

a = (v^2 – v0^2) / 2 d

a = (20^2 – 16^2) / (2 * 36)

a = 2 m/s^2 

The magnitude of the car's acceleration is

[tex]\rm 2\;m/sec^2[/tex]

Step by Step Solution :

Given :

Final Speed, v = 20 m/sec

Initial Speed, u = 16 m/sec

Distance Covered, s = 36 m

Calculation :

We know that,

[tex]\rm v^2-u^2=2as[/tex] --- (1)

Now from equation (1),

[tex]20^2-16^2=2\times a\times 36[/tex]

[tex]a =\dfrac{20^2-16^2}{2\times36}[/tex]

[tex]\rm a = 2 \; m/sec^2[/tex]

The magnitude of the car's acceleration is

[tex]\rm 2\;m/sec^2[/tex]

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