A 5.0 kilogram initially at rest is accelerated by a force of 25 newtons such that it attains 5.0 x 102 joules of kinetic energy. Determine the final speed of the object.

Revolutions per minute are needed for the space station which is at 385m from center of the planet is 1.52 rev/min

Orbital velocity is also called as critical velocity. It is minimum velocity must be given to the satellite or the body, so that it can revolve around the planet. i.e. orbital velocity is minimum velocity of body to revolve in stable orbit around a planet.

Orbital velocity is given by,

[tex]v_{c} = \sqrt{\frac{GM}{R+h}}[/tex] where  G = Gravitational constant (6.673×10⁻¹¹ Nm²/kg²

                               M = Mass of the planet

                               R = Radius of the planet

                                h = height of the object(satellite)

Orbital velocity depends on mass of the planet, radius of the planet and height of the object(satellite). It is independent of mass of the body(satellite).

Given,

Diameter of space station, D = 770m {Radius (R+h) = 385m}

Acceleration due to gravity [tex]g_{h}[/tex]= 9.8 m/s²

our given equation is,

[tex]v_{c} = \sqrt{\frac{GM}{R+h}}[/tex]

[tex]v_{c}^{2} = \frac{GM}{R+h}}[/tex].........1)

we know that v=rω

v²=r²ω²

where r = (R+h) = Radius of planet + height of space station from surface of the planet.

v²=(R+h)²ω²......2)

with equation 2), equation 1 becomes.

(R+h)²ω² = [tex]\frac{GM}{R+h}}[/tex]

ω² = GM÷ (R+h)³.............3)

we know that [tex]g_{h} = \frac{GM}{(R+h)^{2} }[/tex].....4)

equ 3 becomes,

ω² = [tex]\frac{g_{h}}{(R+h)}[/tex]

Putting all values in equation,

ω² = 9.8 ÷ 385

ω² = 0.025454

ω = 0.1595 ≅ 0.16 rad/s

let,

ω = 2πn

n= ω ÷ 2π

n = 0.02547 rev/s

n= 0.02547*60 = 1.52 rev/min

Hence 1.52 rev/min is needed for space station at artificial gravity of 9.8 m/s

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The force required to lift a load of 100 newtons using machine with mechanical advantage of 4 is 25 newtons.

The ratio of output force to the input force is called as mechanical advantage of a machine. In this case, mechanical advantage is given as 4.

As, Mechanical Advantage = Output Force / Input Force

4 = Output Force / Input Force

So, Output Force = Mechanical Advantage x Input Force

Output Force = 4 x Input Force

100 newtons = 4 x Input Force

Input Force = 100 newtons / 4

Input Force = 25 newtons

Therefore, the force required to lift a load of 100 newtons using a machine with a mechanical advantage of 4 is 25 newtons.

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In a genetic cross Rr x rr the percentage of offspring will be round will be 50% since this trait is associated with the recessive genotype (rr).

The recessive genotype in a genetic cross is the proportion of offspring that will be homozygous recessive and therefore will express the recessive trait such as in this case round plants that are produced by the combination of recessive and heterozygous parents.

Therefore, with this data, we can see that recessive genotypes in a genetic cross are generated by the combination of two recessive gametes.

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Since the initial velocity of ball A was 3.3 m/s east and the velocity of ball A after the collision is 6.7 m/s east, the velocity of ball A after the collision is 3.7 m/s east.

Velocity is a vector quantity that measures the rate and direction of an object's motion. It can be determined by measuring the displacement of the object over a certain period of time. Velocity is usually expressed in terms of distance per unit of time, such as meters per second (m/s). In physics, velocity is a fundamental concept that is used to describe the motion of objects.

The velocity of ball A after the collision can be calculated using the conservation of momentum equation. The equation is as follows:

m1v1 + m2v2 = m1v1' + m2v2'

where m1 is the mass of ball A, v1 is the initial velocity of ball A, m2 is the mass of ball B, v2 is the initial velocity of ball B, m1' is the mass of ball A after the collision, v1' is the velocity of ball A after the collision, m2' is the mass of ball B after the collision, and v2' is the velocity of ball B after the collision.

In this problem, the equation is:

(0.62 kg)(3.3 m/s east) + (0.62 kg)(1.6 m/s west) = (0.62 kg)(v1') + (0.62 kg)(1.2 m/s east)

We can solve this equation for v1':

(0.62 kg)(3.3 m/s east) + (0.62 kg)(1.6 m/s west) = (0.62 kg)(v1') + (0.62 kg)(1.2 m/s east)

0.62(3.3 + 1.6) = 0.62v1' + 0.74

4.9 = 0.62v1' + 0.74

4.9 - 0.74 = 0.62v1'

4.16 = 0.62v1'

v1' = 6.7 m/s east

Since the initial velocity of ball A was 3.3 m/s east and the velocity of ball A after the collision is 6.7 m/s east, the velocity of ball A after the collision is 3.7 m/s east.

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Based on the densities of the two liquids, the height of the light liquid in the right arm of the U-tube is 0.203 cm.

Let's first consider the situation before the light liquid is added. At this point, the heavy liquid fills both arms of the U-tube to the same height, h.

The pressure at point A is equal to the pressure at point B

Therefore:

P₀ + ρgh = P₀ + ρgh

where P₀ is the atmospheric pressure, ρ is the density of the heavy liquid, and g is the acceleration due to gravity.

Simplifying this equation, we get:

ρgh = ρgh

Canceling out the ρ and solving for h, we get:

h = h

In other words, the height of the heavy liquid is the same in both arms of the U-tube.

Now let's consider the situation after the light liquid is added to the right arm of the U-tube. We want to find the height, L, of the light liquid in the right arm.

Since the pressure at any two points in a connected vessel is the same, the pressure at point B (the top of the heavy liquid in the right arm) must be equal to the pressure at point C (the top of the light liquid in the right arm).

Therefore, we can write:

P₀ + ρgh = P₀ + ρg(L+h)

where L is the height of the light liquid in the right arm.

Simplifying this equation, we get:

ρgh = ρgL + ρgh

Canceling out the ρgh and solving for L, we get:

L = (ρ/ρ₀)h

where ρ₀ is the density of the light liquid.

Substituting the given values, we get:

L = (0.92 g/cm³ / 13 g/cm³)h

L = 0.070769h

Now we need to find h. We can use the fact that the volume of the heavy liquid in the left arm is equal to the volume of the heavy liquid plus the light liquid in the right arm.

The volume of the heavy liquid in the left arm is:

V₁ = Ah = (13.2 cm²)(h cm)

V₁ = 13.2h cm³

The volume of the heavy liquid plus the light liquid in the right arm is:

V₂ = A(L+h) = (2.11 cm²)(L+h cm)

V₂ = 2.11(L+h) cm³

Since these volumes are equal, we can set them equal to each other and solve for h:

13.2h = 2.11(L+h)

13.2h = 2.11L + 2.11h

11.09h = 2.11L

h = (2.11/11.09)L

Substituting this into our expression for L, we get:

L = 0.070769(2.11/11.09)L

L = 0.01345L

L = 0.01444h

Substituting the given value for the density of the heavy liquid, we get:

L = 0.01444h = 0.01444(13 g/cm³)/(0.92 g/cm³)

L = 0.203 cm

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Complete question:

A heavy liquid with a density 13 g/cm³ is poured into a U-tube as shown in the left- hand figure below. The left-hand arm of the tube has a cross-sectional area of 13.2 cm², and the right-hand arm has a cross-sectional area of 2.11 cm². A quantity of 90.2 g of a light liquid with a density 0.92 g/cm³ is then poured into the right-hand arm as shown in the right-hand figure below.

Determine the height L of the light liquid in the column in the right arm of the U-tube, as shown in the second figure above. Answer in units of cm.