what changes when unbalanced forces are applied to a moving bike along the direction its moving?
A. inertia
B.friction
C. mass
D.velocity

Answers

Answer 1

Answer:

D.velocity

Explanation:

When unbalanced forces are applied to a moving bike along the direction its moving, the velocity of the bike changes.

Unbalanced forces causes a body to accelerate. Acceleration is the rate of change of velocity with time.

When unbalanced forces acts on a body, the velocity changes.

An unbalanced force is a force whose resultant is not zero. Such a force causes a body to change motion and then acceleration. For a body to accelerate, the velocity must change per unit of time.

Related Questions

Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?

a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.

Answers

Energy from the sun is important so the correct answer is c
Explanation

An astronaut with a mass of 75 kg stands on Mars (gravity = 3.7 m/s2). How much does the astronaut weigh?

Answers

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

A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?

A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done​

Answers

Answer:

Explanation:

C 200÷100=2

Output ÷ Input= MA

When an object with an electric charge of is from an object with an electric charge of , the force between them has a strength of . Calculate the strength of the force between the two objects if they are apart. Round your answer to significant digits.

Answers

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

When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?

Answers

Answer:

I think the answer is 19.8 potential energy

Explanation:

NONE.

A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?

Answers

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

According to Newton’s law of universal gravitation, in which of the following situations does the gravitational attraction between two bodies always increase

Answers

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

Which has the most mass?
O The Moon
O A Pencil
O Your teacher.
O Earth

Answers

Answer:

Earth

lol... ....

How does the force of friction for a sliding object vary with the area of contact?

Answers

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.

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

Answers

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

Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do

Answers

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.

what is energy? list the three mechanical forms of energy and their associated equations. ​

Answers

Answer:

mm hope that helps

Explanation:

What are the 3 forms of energy?

Forms of energy

Potential energy.

Kinetic energy.

Your __________ environment includes the people you spend time with.

Answers

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.

A Geiger–Müller tube can be used to detect ionising radiation. The tube is filled with low pressure gas. How does this detect ionising radiation

Answers

Explanation:

the gas will be hit by high energy particles as the radioactive decay happens and the decaying nuclei are breaking apart. these particles energize the gas and cause electrons to move around the orbitals. the flow of electrons is detected by the sensor.

When a rattlesnake strikes, its head accelerates from rest to a speed of 22 m/s in 0.48 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast? Answer in units of W.

Answers

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.

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. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?

Answers

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.

A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.
a. Determine the value of the frictional force exerted on the car.

b. Determine the value of the coefficient of friction between the tires and the road.

Answers

(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

According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.

Answers

I think the answers are a and c

The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.

Show work

Answers

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

a 14n force is applied for 0.33 seconds, calculate the impulse

Answers

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

Force=14NTime=0.33s

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

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

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

An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (average). Calculate magnitude of the plane's velocity, relative to the ground.

Answers

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg       .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î  + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 )  î  (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = [tex]\sqrt{(42.025^2+(-491.92)^2)} km/h[/tex]

the magnitude of Vpg = 493.711 km/h

Which of the following is a concern about the use of wind energy?
A. Wind farms produce a large amount of air and water pollution.
B. Wind farms may pose a threat to birds and bats.
O
C. Wind farms produce radioactive waste.
D. Wind farms often change the local weather patterns.
SUBMIT

Answers

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

What is wind energy?

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.

Which is a concern about the use of wind energy?

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..

To know more about wind energy, click here

https://brainly.com/question/29293441

#SPJ1

Suppose you drop a superball of massMand a marble of mass m(both treated as point masses) from a heighthwith the marble just on top of the superball. A superbowl has essentially elastic collisions with hard objects. Thesuperball hits the floor, rebounds and then collides with the marble. How highdoes the marble go if all the motion is vertical

Answers

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

How can force-time and force-displacement graphs be used to find the impulse or work done?


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

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

C. Gradient of force-time graph & Gradient of force-displacement graph

D. Gradient of force-displacement graph & Gradient of force-time graph

Answers

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.

A strong man and a weak man are trying to carry a ladder . How should they carry it in such a way that the weak person feels less weight of the ladder

Answers

Answer:

The strong person should carry the ladder at the  front end and the weak person should carry it at the back end.

Explanation:

this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.

However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object

                                                                                                                               

I Hope it help you

Answer:

  weak person: near the end of the ladder

  strong person: nearer the center of the ladder

Explanation:

The closer the strong person is to the center of mass of the ladder, the greater the fraction of the ladder's weight that person will carry. The weak person will feel less weight if the strong person is nearer to the center of mass.

For example, if the strong person is 3/4 of the ladder's length from the end where the weak person is, then the strong person will carry twice the weight the weak person is having to carry. (Assuming the mass is uniformly distributed.)

Which of these represent approaches to psychological science? (Choose every correct answer.)
Behavioral
Chemical
Investigative
Metaphysical
Sociocultural
Cognitive
Humanistic

Answers

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:

The conductMental processes of individuals and human groups in different situations,Human experience

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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A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)

Answers

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.

What is Displacement?

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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One disadvantage to experimental research is that experimental conditions do not always reflect reality.


Please select the best answer from the choices provided

T
F

Answers

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

A tank, in the shape of a cube with each side measuring 2 meters, is half-full of water. Above the water is air that is pressurized to 5 kPa. Calculate the resultant force on one side of the tank due to the air and water (ignore atmospheric pressure) and determine the height of this force above the bottom of the tank.

Answers

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 .

16+109=................

Answers

Answer:

=125 tysm for points

Explanation:

Answer: 125

Explanation: I good at addition

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