Calculate the amount Electrical energy consumed by Electrical motor rated 1.5hp if operated for 7 hours?​

Answer:

When the mass increases or when distance between the bodies reduces

Explanation:

According to Newton's law of universal gravitation, the gravitational attraction between two bodies always increase if the mass increases and the distance between the bodies reduces.

The law of universal gravitation states that "the gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

Mathematically;

      Fg  = [tex]\frac{G m1 m2}{r^{2} }[/tex]  

G is the universal gravitation constant

m is the mass

r is the distance

Answer:

Unbalanced Forces.

Explanation:

Answer:

Unbalanced

Explanation:

Unbalanced forces are not equal and they always cause the motion of an object to change the speed and or direction of the object.

Balanced forces are equal and will not cause the motion of an object to change.

From Newton's law of inertia "an object will continue in its state of rest or of uniform motion unless acted upon by an external force".

The net force for a body with unbalanced forces is greater than zero.

For a body with balanced forces, the net force is zero

Answer:

P = 85.72 W

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 22 m/s

Time, t = 0.48 s

Mass, m = 170 g = 0.17 kg

Let a be the acceleration of the rattlesnake.

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]

Let x is the displacement of a rattlesnake. It can be given by :

[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]

The power of the rattlesnake is given by :

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]

So, the power of the rattlesnake is 85.72 W.

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

Answer:

Explanation:

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more .

Hence displacement is more in the downward slopping.

Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more in the downward slopping.

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Answer:

=125 tysm for points

Explanation:

Answer: 125

Explanation: I good at addition

Answer:

social

Explanation:

Your social environment includes the people you spend time with – your family, your friends, classmates and other people in your community.  Your social environment is healthier when you choose friends who show concern for their own health and yours.

Answer:

mm hope that helps

Explanation:

What are the 3 forms of energy?

Forms of energy

Potential energy.

Kinetic energy.

Answer:

12J

Explanation:

Given parameters:

Mass  = 1.5kg

Impulse  = 6kgm/s

 let us start first by find the velocity with which this body moves;

       Impulse  = mass x velocity

         Velocity  = Impulse /  mass  = 6/ 1.5  = 4m/s

Initial velocity  = 0m/s

Unknown:

Resulting kinetic energy  = ?

Solution:

To solve this problem use the formula below:

      K.E  = [tex]\frac{1}{2}[/tex]  m (v  - u)²  

m is the mass

v is the final velocity

u is the initial velocity

  So;

        K.E  =  [tex]\frac{1}{2}[/tex]  x 1.5 x (4 - 0)²  

  K.E  = 1.5 x 8  = 12J

Use of wind energy may pose a threat to birds and bats.

The power of wind is harnessed to generate renewable energy through the use of wind turbines. These turbines are designed to capture the kinetic energy of the wind and convert it into electricity by rotating their blades. This process involves the movement of turbines, which in turn drives a generator that produces electricity. Wind energy is an eco-friendly and sustainable form of energy that does not emit any harmful gases or pollutants into the atmosphere. As countries worldwide seek to reduce their dependence on fossil fuels and transition towards a low-carbon economy, wind energy is gaining popularity as a clean source of electricity.

The implementation of wind energy has raised concerns about its impact on wildlife, specifically birds and bats. The use of wind turbines can lead to collisions and habitat disruptions for these animals. Despite this, studies have shown that wind turbines cause fewer fatalities for birds and bats in comparison to other human-related sources such as buildings, power lines, and vehicles. To mitigate the impact on wildlife, measures like radar technology to detect animal movement and shutting down turbines during migration periods are being employed..

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The question is incomplete, the complete question is;

When an object with an electric charge of −7.0μC is 5.0cm from an object with an electric charge of 4.0μC, the force between them has a strength of 100.7N. Calculate the strength of the force between the two objects if they are 1.7cm apart. Round your answer to 2 significant digits

Answer:

865.1 N

Explanation:

F1 = Kq1q2/r1^2 ---------1

F2 = Kq1q2/r2^2 -------2

We have that;

r1 = 5cm

r2 =1.7 cm

F1 = 100.7 N

Comparing equations 1 and 2

F2 = F1r1^2/r2^2

F2 = 100.7N[(5cm)^2/(1.7cm)^2]

F2= 865.1 N

Answer:

Explanation:

A ) At constant volume :

ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .

For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J

ΔEint = .7 x 12.45 x ( 564 - 300 )

= 2300.76 J .

W₁ = 0 because volume is constant so work done by gas is zero .

Q₁ = ΔEint = 2300.76 J

B )

At constant pressure

Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .

For monoatomic gas ,

Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J

Q₂ = .7 x 20.75 x 264 J

= 3834.6 J

W₂ = work done by gas

= PΔV = nRΔT

= .8  x 8.3 x 264

= 1752.96 J

ΔEint = Q₂ - W₂

= 3834.6 - 1752.96

= 2081.64 J.

 ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as

a)

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Q2= 3834.6 J

W2= 1752.96 J

dE2= 2081.64 J.  

a)

Generally, the equation for the Constant Volume   is mathematically given as

dE = n Cv x dT

Where

Cv = 3/2

R = 1.5 x 8.3 J

R= 12.45 J

Therefore

dE = 0.7 x 12.45 x ( 564 - 300 )

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Generally

Q2 = n Cp dT

Where

Cp = 5/2

R = 2.5 x 8.3 J

R= 20.75 J

Hemce

Q2 = 0.7 x 20.75 x 264 J

Q2= 3834.6 J

For Work done

W=PdV

W= nRdT

Therefore

W= 0.8  x 8.3 x 264

W2= 1752.96 J

Hence

dE = Q₂ - W₂

dE= 3834.6 - 1752.96

dE2= 2081.64 J.  

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Answer:

Explanation:

The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .

Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .

Pressure of water column  = .5 x 10³ x 9.8 = 4.9 k Pa .

Air is pressurized to 5 kPa so

resultant pressure  on one side of the tank due to the air and water

= 4.9 + 5 kPa = 9.9 kPa .

Total force on one face = pressure x area of one face under water

= 9.9 x 10³ x .5 x 2²

= 19.8 kN .

Answer:

The force and texture of an object matter a lot.

Explanation:

If you were to try and run up a glass hill, could you? You maybe could, but it would be harder to than up carpet.

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

Answer:

4.62 N-s

Explanation:

recall that the formula for impulse is given by

Impulse = Force x change in time

in our case, we are given

Force = 14 N

change in time = 0.33s

Simply substituting the above into the equation for impulse, we get

Impulse = Force x change in time

Impulse = 14 x 0.33

= 4.62 N-s

[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]

[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]

[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]

Answer:

I think the answer is 19.8 potential energy

Explanation:

NONE.

Answer:

WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).

Erosion is represented by the scenario (As the waves recede, they carry the sediment away).

Explanation:

A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.

Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.

Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.

Answer:

cognitive, humanistic, behavioral, sociocultural

Explanation:

Behavioral, sociocultural, cognitive, and humanistic are approaches to psychological science.

Psychology is a term to refer to the discipline that focuses on the study of various topics related to human thought such as:

Due to the above, several subdisciplines have emerged that focus on the study of each of the topics. For example:

Behavioral psychology: focused on the study of human behavior.

Sociocultural psychology: focused on the study of human behavior and thought in different social situations.

Cognitive psychology: focused on mental processes related to learning.

Humanistic psychology: focused on the study of human thought from a comprehensive approach.

According to the above, options A, E, F, and G are correct because they mention different sub-disciplines of psychology while the other options mention terms that are not related to sub-disciplines or psychological sciences.

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Answer:

6.86 m/s

Explanation:

The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.

I suppose that the only force, in this case, is the gravitational force acting on the fish.

Then the gravitational equation of the fish will be:

a(t) = -9.8m/s^2

For the velocity equation we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity of the fish and is what we want to find.

For the position equation we need to integrate over time again to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0

p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m

Then the equation is:

p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t

p(t) = (-4.9 m/s^2)*t^2 + v0*t

We know that the maximum height is 2.4m

The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:

v(t) = (-9.8m/s^2)*t + v0 = 0

      t = v0/9.8m/s^2

Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m

2.4m = p( v0/9.8m/s^2) =  (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)

2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))

2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)

2.4m = 0.5*v0^2/(9.8m/s^2)

2.4m/0.5 = v0^2/(9.8m/s^2)

4.8m*(9.8m/s^2) = v0^2

√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s

Answer:

Earth

lol... ....

(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude a of

a = v ² / R

where

v = tangential speed

R = radius of the curve

so that

a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

∑ F = Fs = m a

where

Fs = magnitude of static friction

m = mass of the car

Then

Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N

(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

N - W = 0

where

N = magnitude of normal force

W = weight

so that

N = W = m g = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:

Fs = µ N

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58

Answer:

 h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Explanation:

Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’

let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy

starting point. Starting point

         Em₀ = U = m g h

final point. Just before the crash

         Em_f = K = ½ m v²

as there is no friction the mechanical energy is conserved

         Em₀ = Em_f

         mg h = ½ m v²

         v = √2gh

this speed is the same for the two bodies.

Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment

initial instant. Just when the superball starts contacting the ground

      p₀ = M v

this velocity is negative because it points down

final instant. Just as the superball comes up from the floor

      p_f = M v '

the other body does not move

      p₀ = p_f

     - m v = M v '

       v ’= -v

Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.

Let's use the subscript 1 for the marble and the subscript 2 for the superball

Third part. The superball and the marble collide

the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved

initial instant. Moment of shock

        p₀ = M [tex]v_{1'}[/tex]+ m v_2

final instant. When the marble shoots out.

        P_f = Mv_{1f'}+ m v_{2f}

        p₀ = p_

        M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}

        M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})

in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved

       K₀ = K_f

       ½ M v_{1'}² + m v₂² = M v_{1f'}²  + ½ m v_{2f}²

        M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)

Let's set the relation  (a + b) (a-b) = a² - b²

      M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

let's write our two equations

           M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f})                 (1)

           M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

if we divide these two expressions

           (v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )

we substitute this result in equation 1 and solve

          v_{1f'}= (v₂ + v_{2f}) - v_{1'}

          M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})

           -M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}

          -M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}

          v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}

          v_{2f} = - [tex]( \frac{ M-m}{M +m } )[/tex]) v₂ + 2 [tex](\frac{M}{M+m})[/tex] v_{1'}

now we can substitute the velocity values ​​found in the first two parts

          [tex]v_{2f}[/tex] = - ( \frac{ M-m}{M +m  } ) √2gh + 2(\frac{M}{M+m}) √2gh

we simplify

          v_{2f} = [( \frac{ M-m}{M +m  } ) + 2 (\frac{M}{M+m})] [tex]\sqrt{2gh}[/tex]

let's call the quantity in brackets that only depends on the masses

          A = ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]

           v_{2f}= A \sqrt{2gh}

in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball

finally with the conservation of energy we find the height that the marble reaches

Starting point

          Emo = K = ½ mv_{2f}²

Final point

          Emf = U = m g h'

          Em₀ = Em_f

          ½ m v_{2f}² = m g h ’

          h ’= ½ v_{2f}² / g

         h ’= ½ (A \sqrt{2gh})² / g

         h ’= A² h

         h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Answer:

A. Area under force-time graph & Area under force-displacement graph

Explanation:

To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.

 Impulse  = Force x time

 Work done  = Force x displacement

When we plot a graph of force and time, the area under it is the impulse.

When a graph of force and displacement is plotted, the area under is the work done.

Answer:

y = -2.69 m

the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Explanation:

his problem must be solved with the missile launch equations.

Let's start by looking for the jumper's initial velocity

          R = v₀² sin 2θ / g

for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m

          v₀² = R g / sin 2te

          v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]

          v₀ = 8.80 m / s

Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed

          sin 45 = [tex]v_{oy}[/tex] /v₀

          cos 45 = v₀ₓ / v₀

         v_{oy} = v₀ sin 45

          v₀ₓ = v₀ cos 45

          v_{oy} = 8.8 sin 45 = 6.22 m / s

          v₀ₓ = 8.8 cos 45 = 6.22 m / s

now let's calculate the sato with these speeds

           x = [tex]v_{ox}[/tex] t

the minimum jump is x = 10 m

           t = x / v₀ₓ

           t = 10 / 6.22

           t = 1.61 s

let's find the vertical distance for this time

           y = v_{oy} t - ½ g t²

where zero is placed on the jump building

           y = 6.22 1.61 - ½ 9.8 1.61²

            y = -2.69 m

Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Answer:

277.5 N.

Explanation:

From the question given above, the following data were obtained:

Mass (m) of astronaut = 75 kg

Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²

Weight of astronaut on Mar (Wₘ) =.?

Weight of an object is simply defined by the following equation:

Weight (W) = mass × acceleration due to gravity (g)

W = m × g

With the above formula, we can obtain the weight of the astronaut on Mar as follow:

Mass (m) of astronaut = 75 kg

Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²

Weight of astronaut on Mar (Wₘ) =.?

Wₘ = m × gₘ

Wₘ = 75 × 3.7

Wₘ = 277.5 N

Thus, the weight of the astronaut on Mar is 277.5 N