A car accelerates from rest to 35m/s in a distance of 88 m. What was its acceleration in m/s2, assumed constant?

a. The magnitude and direction of the centripetal acceleration and force when he is spinning at constant angular velocity is 8.72 m/s^2 and 654.0 N respectively.

b.  The astronaut is experiencing 0.89 g when moving at constant angular velocity.

c. The torque that is needed to bring the centrifuge to a stop 6875 Nm.

Angular velocity is described as a pseudovector representation of how fast the angular position or orientation of an object changes with time.

The magnitude of the centripetal acceleration and force, we will use the formula: a = v^2 / r, where v is the tangential velocity and r is the radius of the centrifuge.

a = (2pi20m15.02pi/60)^2 / 20m = 8.72 m/s^2

To calculate the force, we will use the formula

F_ = ma, where m is the mass of the astronaut, 75.0 k

F_ = 75.0 kg * 8.72 m/s^2 = 654.0 N

b. To calculate  the number of g's the astronaut is experiencing when moving at constant angular velocity, we will divide the centripetal acceleration by the acceleration due to gravity, 9.8 m/s^2

8.72 m/s^2 / 9.8 m/s^2 = 0.89 g

c.

Torque = I * alpha, where I is the moment of inertia and alpha is the angular acceleration.

I = (1/2) * 5500.0 kg * 20.0m^2 = 55000 kgm^2

The angular acceleration can be found using the formula

Alpha = (change in angular velocity) / (change in time)

The change in angular velocity is 15.0 rpm - 0 rpm = 15.0 rpm and the change in time is 2.0 minutes = 120 seconds

alpha = 15.0 rpm / 120 s = 0.125 rad/s^2

Torque = 55000 kgm^2 * 0.125 rad/s^2 = 6875 Nm

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The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.

option C

Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.

So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.

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Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

The moment of inertia of the second disk will be  \(I=3.58\ kg-m^2\)

The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.

here it is given that

MOI of disk one   \(I_1=3.4\ kg-m^2\)

Angular velocity  \(w_1=6.1\ \frac{rad}{s}\)

Angular velocity of disk two  \(w=1.8\ \frac{rad}{s}\)

MOI of the disk two \(I=?\)

The final angular velocity \(w_f= 1.8\ \frac{rad}{sec}\)

Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.

\(I_1w_1+I_2w_2=(I_1+I_2)w_f\)

Now put the values in the formula

\((3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8\)

\(I_2=3.58\ kg-m^2\)

Thus the moment of inertia of the second disk will be  \(I=3.58\ kg-m^2\)

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Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

\(T_{simple\) = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / \(I\) ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and \(I\)  is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

\(I\) = \(\frac{1}{3}\)mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/\(I\)) OR T = 2π√(\(I\)/mgL)

so we can use \(I\) = \(\frac{1}{3}\)mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =  \(\frac{1}{2}\)D.

now, substituting these equations, the period becomes;

T = 2π/√(\(I\)/mgL) OR T = \(2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }\) OR T = 2π√(2D/3g )  ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω\(_{simple\) = 2π/\(T_{simple\) OR  ω\(_{simple\) = √(g/D) OR  ω\(_{simple\) = 2π√( D/g )  

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × \(T_{simple\)

we substitute in value of \(T_{simple\)

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Explanation: I had a question similar to this, Hope this helps!

The gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

The gravitational force between two objects depends on their masses and the distance between them. This force is given by the formula:

F = (G × m₁ × m₂) / r² where F is the gravitational force, m₁ and m₂ are the masses of the two objects, r is the distance between them, and G is the gravitational constant, which has a value of 6.7 × 10⁻¹¹ N m²/kg².

Using this formula, we can find the gravitational force that the moon has on a person with a mass of 50 kg.

The mass of the moon is 7 × 10²² kg, and the distance between the moon and the person is 380,000,000 m.

Therefore, we have:

m₁ = 50 kg

m₂ = 7 × 10²² kg

r = 380,000,000 m

G = 6.7 × 10⁻¹¹ N m²/kg²

Substituting these values into the formula, we get:

F = (G × m₁ × m₂) / r²

F = (6.7 × 10⁻¹¹ × 50 kg × 7 × 10²² kg) / (380,000,000 m)²

F = 1.15 N

Therefore, the gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

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Answer:

    \(\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}\)

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      \(a_A = n \ a_B\)  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = \(\sqrt{L_1^2 + L_2^2}\)  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

Answer:

180

Explanation:

1) E=F*L, where E - energy of the baseball, F - the required force, L - backward moving (0.35m);

2) E=mV²/2, where E - energy of the baseball, m - the mass of the baseball (0.14kg), V - the velocity of the baseball (30m/s).

3) if E=F*L and E=mV²/2, then F*L=mV²/2, from which

\(F=\frac{mV^2}{2L};\)

4) according to the last formula

\(F=\frac{0.14*900}{2*0.35}=\frac{126}{0.7}=180(N).\)

The force exerted by the ball on the glove is 180 Newtons.

In mechanics, a force is any action that has the potential to change, maintain, or deform a body's motion. The three principles of motion outlined by Isaac Newton in his Principia Mathematica are frequently used to illustrate the concept of force (1687).

Newton's first law states that a body at rest or moving uniformly in a straight line will stay in that state until a force is applied to it. According to the second law, a body will accelerate (change in velocity) in the direction of any external force acting on it.

Given:

A baseball (m = 140 g) traveling at 30 m/s moves a fielder's glove backward 35 cm when the ball is caught,

Calculate the value of force as shown below,

Force = m × V² / 2L

Force = 0.140 × 30² / 2 × 0.35

Force = 0.140 × 900 / 0.70

Force = 126 / 0.7

Force = 180 N

Thus, the force exerted by the ball is 180 Newtons.

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The two towers' heights are separated by 201.6 metr

The amount of space between two objects or points is expressed as distance. Several methods, such as linear, angular, and temporal distance, are available to measure it. In physics, math, and other sciences, distance is an important notion. It is employed to describe the size of physical phenomena, like the force of gravity or the speed of light. It can also be used to estimate the distance between two points in time or space. Distance can also refer to the spacing between two locations or people in everyday life.

To estimate the distance between the two towers' heights,we must use the Pythagorean Theorem. The square of the hypotenuse, or side opposite the right angle, is equal to the sum of the squares of the other two sides, according to the Pythagorean Theorem. The heights of the two towers are represented by the other two sides, and the hypotenuse in this case represents the distance between the tops (150 m and 136 m).

This equation can be written as a2 + b2 = c2, where a = 150 m, b = 136 m, and c = the separation between the summits of the two towers. By first squaring both a and b, and then subtracting the result from both sides of the equation, we can find c:

150² + 136² = c²\s22500 + 18496 = c²\s40996 = c²\sc

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a ) The normal force on the crate in terms of the F = 921.2 - 0.5 F N

b ) The minimum force the man would have to apply to pull the crate up the ramp = 533.44 N

W = m g

m = 100 kg

g = 9.8 m / s²

W = 100 * 9.8

W = 980 N

Resolving the force of gravity into its horizontal and vertical components,

Wx = W sin θ

Wx = 980 * sin 20°

Wx = 333.2 N

Wy = W cos θ

Wy = 980 * cos 20°

Wy = 921.2 N

Resolving F into its vertical and horizontal components,

Fx = F cos θ

Fx = F cos 30°

Fx = 0.87 F N

Fy = F sin θ

Fy = F sin 30°

Fy = 0.5 F N

Since there is no acceleration in y-direction,

N - Wy + Fy = 0

N = 921.2 - 0.5 F

Ff = μ N

If the crate is being pulled with the minimum force, velocity would be constant. So acceleration will be zero.

∑ \(F_{x}\) = 0

Fx - Ff - Wx = 0

0.87 F - 0.2 ( 921.2 - 0.5 F ) - 333.2 = 0

0.87 F - 184.24 + 0.1 F - 333.2 = 0

0.97 F = 517.44

F = 533.44 N

Therefore,

a ) The normal force on the crate in terms of the F = 921.2 - 0.5 F N

b ) The minimum force the man would have to apply to pull the crate up the ramp = 533.44 N

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Answer:

The decay constant, or "lambda" (λ), is the rate at which a radioactive isotope decays. It is usually measured in units of inverse time, such as seconds. In this case, the decay constant can be calculated as follows:

16:42

λ = (ln(2)/3.57 x 106) x (5.78 x 1017) = 0.

Explanation:

Answer: The capacitor is fully charged when the voltage of the power supply is equal to that at the capacitor terminals. This is called capacitor charging; and the charging phase is over when current stops flowing through the electrical circuit.

AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:

Explanation:

Initial acceleration is the acceleration of an object at a specific moment in time, typically at the start of a motion. It is the acceleration of an object at the beginning of its motion or when it is first subjected to a force.

Acceleration is a vector quantity that describes the rate at which an object's velocity changes with respect to time. It is the measure of how quickly the object's speed and/or direction is changing. An object can have varying accelerations depending on the forces acting on it, such as gravity or friction.The initial acceleration can be calculated using the formula a = v^2/r, where v is the initial velocity and r is the radius of circular path.

To solve this problem, use the conservation of energy and Newton's second law.

First, calculate the height of point B above point A.

Use the fact that the vertical component of the motorcycle's velocity is zero at the top of the circle, so the kinetic energy is entirely due to the horizontal component of velocity. Therefore, at point B, the kinetic energy of the motorcycle is equal to the potential energy it had at point A:

\(mgh = 1/2 mv^2\)

where m is the mass of the motorcycle, g is the acceleration due to gravity, h is the height of point B above point A, and v is the speed of the motorcycle at point B.

Cancel out the mass of the motorcycle, and use the given values to solve for h:

\(gh = 1/2 v^2 - 1/2 vA^2\)

\(h = (1/2 v^2 - 1/2 vA^2) / g\)

\(h = (1/2 (v^2 - vA^2)) / g\)

\(h = (1/2 ((0.04 s) t)^2 - 2^2) / 32.2 ft/s^2\)

\(h = (0.0008 t^2 - 4) / 32.2 ft\)

Next, use the fact that the motorcycle's acceleration is directed towards the centre of the circle, and has a magnitude of:

\(a = v^2 / r\)

where r is the radius of the circle. At point A, the velocity is purely horizontal, so the initial acceleration is:

\(aA = vA^2 / r\)

Use the fact that the acceleration is given by:

\(a = d(˙v)/dt\)

where ˙v is the rate of change of the velocity with respect to time. Integrating this equation gives:

v - vA = ∫a dt

v = vA + ∫a dt

Since the acceleration is constant, substitute the expression  derived for aA and integrate over the time it takes the motorcycle to travel from A to B. We can use the fact that the distance traveled along the circle is equal to the height difference h we calculated earlier, so the time it takes to travel from A to B is:

\(t = sqrt(2h / g)\)

\(t = sqrt((0.0008 t^2 - 4) / 16.1)\)

Squaring both sides and rearranging, we get a quadratic equation in t^2:

\(t^4 - 27.3t^2 + 674.5 = 0\)

Solving for t^2 using the quadratic formula, we get:

\(t^2 = 13.4 or t^2 = 50.3\)

Since the time cannot be negative, take the positive root:

t = 3.66 s

Substituting this value into the expression for the velocity,

\(v = 2 ft/s + (0.04 s/ft/s^2)(3.66 s)\)

\(v = 2.1464 ft/s\)

Therefore, the magnitude of the motorcycle's velocity when it reaches point B is approximately 2.15 ft/s. The initial acceleration is:

\(aA = vA^2 / r = (2 ft/s)^2 / 16 ft = 0.25 ft/s^2\)

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A motorcyclist is traveling along a vertical circular path. At point A, the motorcyclist has an initial velocity of vA = 2 ft/s and an initial position of s = 0. The motorcyclist increases their speed along the path at a rate of ˙v = (0.04s) ft/s^2, where s is in feet. Determine the magnitude of the motorcyclist's velocity when they reach point B. Also, what is the motorcyclist's initial acceleration at point A?

The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.

Given:

Object height (h) = 5 cm

Focal length of the convex lens (f) = 10 cm

Object distance (u) = 15 cm (positive since it's on the same side as the incident light)

To determine the nature, position, and size of the image, we can use the lens formula:

1/f = 1/v - 1/u

Substituting the given values:

1/10 = 1/v - 1/15

To simplify the equation, we find the common denominator:

3v - 2v = 2v/3

Simplifying further:

v = 30 cm

The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.

To find the magnification (M), we use the formula:

M = -v/u

Substituting the values:

M = -30 / 15 = -2

The magnification is -2, indicating that the image is inverted and twice the size of the object.

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A. The total momentum of the system after they push off is zero.

B. m1v1=m2v2

v2=m1v1/m2

v2=80*3.0/32

v2=7.5 m/s

in fact

m1v1=80*3= 240 kgm/s

m2v2= 32*7.5= 240 kgm/s.

A penny board is a type of skateboard characterized by a short, narrow deck made of plastic. If you're on the short side, consider choosing a smaller deck. Smaller decks are easier to operate and control, lighter, and more portable. If you are taller, you should choose a larger deck.

The larger deck gives more stability and power when skating. A big part of being a skater girl is looking like that which means wearing skate-branded shoes and clothes. Other skaters tend to wear these clothes cut to allow the skater's movement. Good skatesHer brands include Vans DC Nike especially basketball lines and Etnies I have.

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Between Kinetic energy and speed: The relationship is quadratic or parabolic. According to the kinetic energy formula, KE = 1/2 mv^2, the kinetic energy is proportional to the square of the speed.

Between potential energy and mass: The relationship is linear. The potential energy is directly proportional to the mass. In simple cases, the potential energy is given by PE = mgh, where mass (m) and height (h) are directly proportional.

Between pressure and volume: The relationship is inverse. According to Boyle's law, the pressure and volume of a gas are inversely proportional when temperature is constant. Mathematically, P1V1 = P2V2.

Between pressure and temperature: The relationship is linear. According to Charles's law, the pressure and temperature of a gas are directly proportional when volume is constant. Mathematically, P1/T1 = P2/T2.

Between displacement and velocity: The relationship is linear. Velocity is the rate of change of displacement with respect to time, so the two are directly proportional.

Explanation:

resultant motion = √x² + y²

this is using Pythagoras theorem whereas the x and y components can be used to construct a right angle triangle with the hypotenuse = resultant.

hope it helps. :)

Answer:

Play tennis, nothing can train you better for the sport than the sport itself. However, tennis is one of those unique sports that combine nearly all components of fitness including power, agility, speed, flexibility, reaction time, balance, coordination, cardiovascular endurance and muscular endurance.

Explanation:

Answer:

26.833 N

Explanation:

The computation of the resaltant of two forces is shown below:

Given that

Force A = 12N

Force B = 24N

Based on the above information  

Resultant R is

\(=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}\)

=26.833 N

According to the given statement  5.50 *  10¹⁴Hz is the frequency of the incident light.

The photoelectric effect, which happens when light strikes a metal, can release electrons out of its surface. As the electrons that are expelled first from metal are known as emitted electrons, this process is also sometimes referred to as photoemission.

The following equation may be used to determine a photon's energy in terms of frequency:

E = hf

The work function must first be changed from electron volts (eV) to joules (J):

1 eV = 1.602 × 10⁻¹⁹ J

Hence, the work function is:

1.68 eV × 1.602 × 10⁻¹⁹ J/eV = 2.69 × 10⁻¹⁹ J

The emitted photon's kinetic energy is:

E = 9.60 × 10⁻²⁰ J

E = E0 + KE

where KE is the kinetic energy of the released electron and E0 is the work function.

Inputting the values, we obtain:

hf = E0 + KE

hf = 2.69 × 10⁻¹⁹ J + 9.60 × 10⁻²⁰J

hf = 3.65 × 10⁻¹⁹ J

When we solve for f, we obtain:

f = E/h = (3.65 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J s)

f = 5.50 × 10¹⁴ Hz

As a result, the incident light has a frequency of 5.50 *  10¹⁴Hz.

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If a frictionless plane is 10.0 m long and inclined at 36.0°.

We can use conservation of energy to find the distance that the first sled travels up the incline. The potential energy of the sled at the bottom of the incline is zero, and its kinetic energy is:

KE = (1/2)mv^2

where m is the mass of the sled and v is its speed. At the point where the sled stops, all of its kinetic energy has been converted into potential energy, so we can write:

mgh = (1/2)mv^2

where h is the height that the sled has traveled up the incline. Solving for h, we get:

h = (v^2)/(2g)

where g is the acceleration due to gravity. Using the given values, we have:

h = (6.00 m/s)^2 / (2 * 9.81 m/s^2) = 1.83 m

So the first sled travels a distance of 10.0 m - 1.83 m = 8.17 m up the incline.

b. To find the initial speed of the second sled, we can use conservation of energy again. At the top of the incline, the sled has potential energy:

PE = mgh

where h is the height of the incline. As the sled slides down the incline, its potential energy is converted into kinetic energy:

KE = (1/2)mv^2

where v is the speed of the sled at the bottom of the incline. We can equate these two expressions and solve for v:

mgh = (1/2)mv^2

v = sqrt(2gh)

Using the given values, we have:

v = sqrt(2 * 9.81 m/s^2 * 10.0 m * sin(36.0°)) = 12.2 m/s

So the second sled must be released from the top of the incline with an initial speed of 12.2 m/s.

The position of the sled as a function of time is given by:

y = -0.5gt^2 + Vi*t + h

where y is the vertical position of the sled, t is the time, Vi is the initial speed of the sled, and h is the height of the incline. At the bottom of the incline, y = 0, so we can solve for the time it takes for the second sled to reach the bottom:

0 = -0.5gt^2 + Vi*t + h

t = (Vi ± sqrt(Vi^2 - 2gh)) / g

Since we want both sleds to reach the bottom at the same time, we set the time for the first sled to slide down the incline equal to this expression for t and solve for Vi:

t = sqrt(2h/g) = sqrt(2 * 1.83 m / 9.81 m/s^2) = 0.619 s

0 = -0.5gt^2 + Vi*t + h

Vi = (h - 0.5gt^2) / t

Vi = (1.83 m - 0.5 * 9.81 m/s^2 * (0.619 s)^2) / 0.619 s

Vi = 5.68 m/s

So the initial speed of the second sled is about  5.68 m/s

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The student with the correct claim and reasoning is Jane, she, “Since the tension force is greater than the friction force, the block will move to the right at a constant velocity.”

option D is the correct answer.

The direction of motion of the engine block depends on the direction of the net force on the engine  block.

The net force on the block is the sum of all the forces acting on the block and can be calculated as follows;

The sum of the vertical forces;

Fy = FN - FW

where;

The sum of the horizontal forces;

Fx = FT - FF

where;

Thus, based on the net force on the engine block, the engine block will horizontally when the tension force on the engine block is greater than the force of friction. In the given force diagram, the tension force is greater than the force of friction and the block will move right.

The block will not accelerate to right since the tension force and friction force are constant.

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The coefficient of kinetic friction between your tires and the road is determined as 0.4.

The acceleration of the car is determined by applying the following kinematic equation as shown below;

v² = u² - 2as

where;

when the car stops, the final velocity of the car = 0

0 = u² - 2as

2as = u²

a = u² / 2s

a = ( 14² ) / ( 2 x 25 )

a = 3.92 m/s²

The coefficient of kinetic friction is calculated by applying the following equations;

μ = Ff / W

where;

At a constant speed, applied force = force of friction

μ = Ff / W

μ = ( ma ) / ( mg )

μ = a / g

where;

μ = 3.92 / 9.8

μ = 0.4

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Answer:

A

Explanation:

Explanation:

To determine the effort needed to lift a load of 200N, given a velocity ratio of 5 and an efficiency of 80%, we can use the formula:

Efficiency = (Output Work / Input Work) * 100

Efficiency can also be calculated as the ratio of the output force to the input force. In this case, the output force is the load being lifted (200N), and the input force is the effort required.

Given that the velocity ratio is 5, it means that for every 5 units of distance the effort moves, the load moves 1 unit of distance. This implies that the effort is exerted over a greater distance than the load.

Let's denote the effort force as "E" and the distance moved by the effort as "dE." Similarly, the load force is "L," and the distance moved by the load is "dL."

Using the velocity ratio, we have the following relationship:

dE / dL = 5

Now, we can calculate the input work (Wi) and the output work (Wo):

Input Work (Wi) = Effort (E) * Distance moved by the effort (dE)

Output Work (Wo) = Load (L) * Distance moved by the load (dL)

Given that the efficiency is 80%, we can rewrite the formula for efficiency as:

0.80 = (Wo / Wi) * 100

Now, let's solve for the effort (E) using the given values:

Load (L) = 200N

Efficiency = 0.80

Velocity Ratio = 5

First, calculate the output work (Wo):

Wo = Load (L) * Distance moved by the load (dL)

Since the velocity ratio is 5, the distance moved by the load (dL) will be 1/5 of the distance moved by the effort (dE):

dL = (1/5) * dE

Wo = L * (1/5) * dE

Wo = 200N * (1/5) * dE

Wo = 40N * dE

Next, calculate the input work (Wi):

Wi = Effort (E) * Distance moved by the effort (dE)

Wi = E * dE

Now, substitute the values into the efficiency formula:

0.80 = (Wo / Wi) * 100

0.80 = (40N * dE) / (E * dE) * 100

0.80 = 40 / E * 100

0.80 * E = 40

E = 40 / 0.80

E = 50N

Therefore, the effort needed to lift a load of 200N with a velocity ratio of 5 and an efficiency of 80% is 50N.

equal in magnitude but opposite in direction

A)  The resistance is 2000 ohm. B) 6 times the current through the muscle. C) The voltage is 1 volt. D) The resistance is 1000 ohms.

A)

r = resistance of muscle

R = resistance of fat

given that: R = 3r

Rtotal = total resistance = R × r /(R + r) = 3r2 /4r= 0.75 r

V = potential difference = 1 Volts

I = current = 1 mA = 0.001 A

Using Ohm's law:

V = I × R(total)

0.5 = (0.001) × (0.75r)

r = 666.67

R = 3r = 3 x 666.67 = 2000

Hence, The resistance is 2000 ohm.

b)

1/6 times the current through the muscle.

since muscle and fat are in parallel, they have the same voltage across each, hence

i(muscle) × R(muscle) = i(fat) × R(fat)

i(muscle) × R(muscle) = i(fat) × 6 × R(muscle)

i(fat) = (1/6) i(muscle)

Therefore, 6 times the current through the muscle.

c)

since fat and muscle are in parallel

Hence, V(fat) = 1 Volts

Hence, The voltage is 1 volt.

d)

R(arm) = arm resistance = 750

R(fat) = fat resistance

R(muscle) = muscle resistance

given that

R(fat) = 3 R(muscle)

since fat and muscle are in parallel their combination is given as

R(arm) = R(fat) ×  R(muscle) ÷ (R(fat) + R(muscle) )

750 = (3 R(muscle) )R(muscle) ÷ ((3 R(muscle))+ R(muscle) )

750 = 3 R(muscle) /4

R(muscle) = 1000

Therefore, The resistance is 1000 ohms.

To know more about Ohm's law:

https://brainly.com/question/1247379

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