Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
(a) Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon? Express your answer in terms of G, m, and r.
(b) Since the gravitational force between two bodies decreases with distance, the acceleration a_near experienced by a unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration a_far experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general expression for the quantity a_near - a_far. Express your answer in terms of G, m, r, and re.

Answer:

C.) transitioning

Explanation:

Look for factors that will help you determine what type of economy exists in Country C is a transitioning economy.

The transition economy or economy in transition is an economy that is changing from centrally planned to free market. These submit to Liberalism, where market forces are prices, rather than a centrally planned organization.

For these reasons we can say that the economy of country C is a transition economy.

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After 480 years have passed, there would be 62.5 grams of this element remaining in the sample.

The half-life of an element is the time it takes for half of a given sample to decay. In this case, the half-life of the element is 120 years.

After 120 years, half of the original sample would remain, which is 1000/2 = 500 grams. After another 120 years (totaling 240 years), half of the remaining 500 grams would decay, leaving 250 grams. Following the same pattern, after a total of 360 years (three half-lives), half of the remaining 250 grams would decay, leaving 125 grams. Finally, after 480 years (four half-lives), half of the remaining 125 grams would decay, leaving 62.5 grams.

Therefore, after 480 years have passed, there would be 62.5 grams of this element remaining in the sample.

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From conservation of linear momentum, the final velocity is 15.7 m/s and the time taken is 1.6 s

Velocity can be defined as a distance travel in a specific direction per time taken. It is a vector quantity.

Given that 30kg metal ball is dropped from a height of 12.5 m.  

a.  Find the final velocity when the ball  hits the ground.

The maximum K.E of the ball at it final velocity will be equal to its maximum P.E at height 12.5 m. That is,

K.E = P.E

1/2mv² = mgh

Where

Substitute all the parameters

1/2 × 30 × v² = 30 × 9.8 × 12.5

v² = 245

v = √245

v = 15.65 m/s

b. The time it takes for the ball to hit the ground can be found through

h = ut + 1/2gt²

but u = 0

h = 1/2gt²

Substitute all the necessary parameters

12.5 = 1/2 × 9.8 × t²

12.5 = 4.9t²

t² = 12.5/4.9

t² = 2.55

t = √2.55

t = 1.6 s

Therefore, the final velocity when the ball hits the ground is 15.65 m/s and the time it takes for the ball to hit the ground is 1.6 s

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Solution :

Mass of the particle = M

Speed of travel = v

Energy of one photon after the decay which moves in the positive x direction = 233 MeV

Energy of second photon after the decay which moves in the negative x direction = 21 MeV

Therefore, the total energy after the decay is = 233 + 21

                                                                           = 254 MeV

So by the law of conservation of energy, we have :

Total energy before the decay = total energy after decay

So, the total relativistic energy of the particle before its decay = 254 MeV  

Answer:

960.4N

Explanation:

Given parameter:

Mass of the body  = 98kg

Unknown:

Force exerted by gravity  = ?

Solution:

Force exerted by gravity on the body is the weight

  Weight  = force x acceleration due to gravity

Acceleration due to gravity = 9.8m/s²

  Weight  = 98kg x 9.8m/s²

  Weight  = 960.4N

As the temperature is increased, the maximum height that the ball reaches is decreased.

We know that an experiment would have to have an independent variable and a dependent variable. The independent variable is the variable that we would have to change as the reaction is taking place. The dependent variable is variable that is altered as we effect changes in the dependent variable.

Having said that, we can see that in the reaction that we are carrying out here, the temperature of the rubber ball is the independent variable while the maximum height the ball reaches after it bounces is the dependent variable. This would help us to be able to know what happens in the experiment that has been described.

The increase in the temperature would cause the gases to expand and the ball would no longer move very high again.

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As per the given question, the asteroid Icarus has a perihelion distance of 0.19 AU and an orbital eccentricity of 0.83. We have to determine its semimajor axis and aphelion distance from the sun.

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Answer:

I think its the electromagnetic spectrum

Answer:   Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.

Explanation:

Answer: you have to divide and multiple fam

Explanation:

The high reached by the block when it approaches point C, depends of the potential energy at the beginning of the motion.

The potential energy is:

U = m·g·h

due to the path is frictionless, the energy of the system is constant. Then, while the block approaches to point C, all its energy becomes potential energy and this potential energy must be the same that the initial potential energy.

Due to the mass m and g are constant, and U must be the same, then, it is necessary that h, the height, is the same that at the beginning of the motion.

Hence, the hight is 4.5 m (the block does not reache point C) as the block approaches point C.

Answer:

Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.

Hope this helps..

The acceleration of the falling obeject of mass 2.55 kg is  -8.2 m/s². And the right option is d) -8.2 m/s².

This can be defined as the rate of change of velocity.

To calculate the acceleration of the object, we use the formula below.

Formula:

Where:

Note: Taking forces acting upward to be positve and downward force to be negatove,

From the question,

Given:

Substitute these values into equation 2

Hence, The acceleration of the object is -8.2 m/s².

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Answer:

balanced forces

Answer:

it's answer is 117 N

F = Gm1m2/r^2

F = 6.67 * 10 ^-11 * 70 * 7.34 * 10 ^22/(1.71 * 10 ^6)^2

F = 117 N

hope it helps you

Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:

        A) The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

Given parameters

To find

   A)  Apparent weight

   B)  Factor that exceeds the actual weight

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

This exercise we will do them in two parts

1st part. Freefall

We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero

            v = v₀ - g t

Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time

            v = -gt

            v = -9.8 2.0

            v = -19.6 m / s

The negative sign indicates that the velocity is directed downwards

2nd part. Braking movement

We look for the acceleration to stop the body, in this case the final velocity is zero

            \(v_f\) = v + a t

             0 = v + at

             a = \(- \frac{v}{t}\)- v / t

             a = \(-\ \frac{-19.6}{0.5}\)

             a = + 39.2 m / s²

The positive sign indicates that the acceleration is directed upward.

A) Newton's second law states that the force is proportional to the mass and the acceleration of the body

            ∑ F = m a

            F -W = m a

            F = m (a + g)

            F = 65 (39.2 + 9.8)

            F = 3185 N

Body weight is

             W = mg

             W = 65 9.8

             W = 637 N

therefore the apparent weight is

             ΔW = F -W

             ΔW = 3185 - 637

             Δw = 2548 N

B) the weight change factor

             Factor =\(\frac{F}{W}\) F / W

             Factor =\(\frac{3185}{937}\)

             factor = 5

In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:

       A)  The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

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The work done on 85 N boxes of books while carried horizontally 4m across a room is equal to zero. Therefore, option (C) is correct.

Work can be demonstrated as the energy used when a force is exerted to move an object through a definite displacement. The work done defines both the exerted force on the object and the distance moved by the object.

The applied force moves an object in a straight line in the direction of the exerted force over a displacement 'd'.

W= F × d

Where 'F' is the exerted force and 'd' is the displacement and W is work done.

Given, the weight of the box = 80 N

The displacement , d = 4 m

In order for to work to be done the box should move in the direction of exerted force as the books are moving forward direction and the applied force in an upward direction to carry the box.

Therefore, the distance moved in the direction of force is zero so the work done is also equal to zero.

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Answer:

40m

Explanation:

distance = length traveled = 20+20 = 40

Answer:

True

Explanation:

Answer: 4.94cm³

Explanation:

Data;

ρ = 1.59g/cm³

mass = 7.85g

volume = ?

density = mass / volume

ρ = m / v

v = m / ρ

v = 7.85 / 1.59

v = 4.94cm³

The velocity of the air relative to the runner is 5 m/s.

We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;

Velocity of the runner = 4m/s. due west

Velocity of the wind =  3m/s due south

The relative velocity is;

Vr = √(4)^2 + (3)^2

Vr = 5 m/s

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The acceleration due to gravity of the asteroid is 833.85 N.

F = mg

where;

Assuming the acceleration due to gravity of the asteroid = 9.81 m/s²

F = 85 x 9.81

F = 833.85 N

Thus, the acceleration due to gravity of the asteroid is 833.85 N.

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Answer and Explanation:

A.

There would be a net force of 50 N going to the right.

Since both forces are going right, they add onto the force amount.

#teamtrees #PAW (Plant And Water)

The final velocity of the 25.8 g marble after the collision is 16.15 cm/s.

The velocity of the 25.8 g marble after the collision is calculated as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

The final velocity of the 25.8 g marble after the collision is calculated as;

( 25.8 x 21 ) + ( 12.4 x 13.8 ) = ( 12.4 x 23.9 ) + ( 25.8v )

712.92 = 296.36 + 25.8v

25.8v = 416.56

v = 416.56 / 25.8

v = 16.15 cm/s

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Answer:

Incomplete question but the closest ans is

A) The displacement is not zero

Explanation:

For work to be done, the direction of force has to be parallel to or in the same direction as the displacement, this can then be calculated by

\(W=Fs \ cos\)θ

Where θ is the angle between direction of force and direction of displacement

By this equation,

if Force is zero, W is zero

if displacement is zero, W is zero

if force and displacement are perpendicular θ=90, cosθ = 0

So A is the cloest choices

Answer:

The weight of the body in the new planet is 100 newtons.

Explanation:

From Newton's Law of Gravitation we find that gravitational force is directly proportional to mass of the planet and inversely proportional to the square of its radius. From this fact we can build the following relationship:

\(\frac{F_{1}\cdot R_{1}^{2}}{M_{1}} = \frac{F_{2}\cdot R_{2}^{2}}{M_{2}}\) (1)

Where:

\(F_{1}\), \(F_{2}\) - Gravitational force, measured in newtons.

\(M_{1}\), \(M_{2}\) - Mass of planet, measured in kilograms.

\(R_{1}\), \(R_{2}\) - Radius of the planet, measured in meters.

If we know that \(F_{1} = 800\,N\), \(\frac{M_{2}}{M_{1}} = \frac{1}{2}\) and \(\frac{R_{2}}{R_{1}} = 2\), then the expected gravitational force in the new planet is:

\(F_{2} = F_{1}\cdot \left(\frac{M_{2}}{M_{1}} \right)\cdot \left(\frac{R_{1}} {R_{2}}\right)^{2}\)

\(F_{2} = (800\,N)\cdot \left(\frac{1}{2} \right)\cdot \left(\frac{1}{2}\right)^{2}\)

\(F_{2} = 100\,N\)

The weight of the body in the new planet is 100 newtons.

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d -  I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L =  (μI/2π) x (I/2d + I/d)

F₃/L =  (μI/2π) x (3I/2d)

F₃/L =  3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

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