how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy flows.

Answer: 6.22 N(the units have been mentioned wrongly in the question)

Given:

mass of the ball(m)=0.42 Kg

acceleration of the ball(a)=14.8m/s^2

F=mxa

Where m is the mass of the ball.

a is the acceleration of the ball.

F is the force applied on the ball.

F=0.42 X 14.8

F= 6.26 N

Initial distance (D1):75 Km

Time taken to cover this distance= 3 hrs.

Speed= Distance /time

Speed1= 75/3= 25Km/hr

Later the train travels

Distance (D2):66 Km

Time taken to cover this: 1hr

Speed= Distance/time

Speed2= 66/1= 66Km/hr

Average speed= (speed1+ speed 2)/2

Average speed= (25+66)/2

Average speed= 45.5 Km/hr

No, the mechanical energy does not change if it is ideal.

We define mechanical energy as the sum of kinetic energy and potential energy

ME = PE + KE

When the pendulum swings, potential energy is converted to kinetic energy and vise versa. But the sum of both remains constant, which is the mechanical energy.

Force on left crate, right crate and truck we need to draw here

First we will write the equation for left crate

[tex]T_{left} = m*a[/tex]

[tex]T_{left} = 20*a[/tex]

Now similarly for right crate

[tex]T_{right} - T_{left} = 20*a[/tex]

now for truck

[tex]F - T_{right} = 20*a[/tex]

Now we know that F = 200 N

[tex]200 - T_{right} = 20*a[/tex]

now add all three equation

[tex]T_{left} + T_{right} - T_{left} + 200 - T{right} = 20a + 20 a + 20a[/tex]

[tex]200 = 60 a[/tex]

[tex]a = 10/3 m/s^2[/tex]

now we will find all tension using this value of acceleration

[tex]T_{left} = 20 * \frac{10}{3} = \frac{200}{3} N[/tex]

[tex]T_{right} - \frac{200}{3} = 20*\frac{10}{3}[/tex]

[tex]T_{right} = \frac{400}{3} N[/tex]

Answer:

Option C is the correct answer.

Explanation:

  We have velocity of a body = Change in position/ Time.

  Considering first portion of graph,

  Change in position = 30 - 0 = 30 m

  Time = 0.75 hours = 45 minutes = 2700 seconds

   Velocity = 30/2700 = 0.011 m/s

  Considering second portion of graph,

  Change in position = 30 - 30 = 0 m

  Time = 0.5 hours = 30 minutes = 1800 seconds

   Velocity = 0/1800 = 0 m/s

 Considering third portion of graph,

  Change in position = 0 - 30 = -30 m

  Time = 0.75 hours = 45 minutes = 2700 seconds

   Velocity = -30/2700 = -0.011 m/s

So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).

Option C is the correct answer.

Answer:

I had this same question the answer is C 100% sure

Explanation:

plane is flying at an altitude of 70 m

now if an object is dropped from it then time taken by object to drop on ground will be given as

[tex]y = v_i* t + \frac{1}{2}at^2[/tex]

here initial speed in vertical direction must be zero as plane is moving horizontal

given that

y = 70 m

a = 9.8 m/s^2

[tex]70 = 0 + \frac{1}{2}*9.8*t^2[/tex]

[tex]t = 3.77 s[/tex]

now since the plane is moving horizontally with speed v = 44 m/s

so the horizontal distance moved by the object will be

[tex]d = v_x * t[/tex]

[tex]d = 44 * 3.77 [/tex]

[tex]d = 166.3 m[/tex]

so the distance moved by the box is 166.3 m

Answer:

1.7 × 10² m

Explanation:

The movement of the weight can be decomposed in a vertical component and a horizontal component.

The vertical movement is uniformly accelerated motion (constant acceleration) and is the one that we will use to find the time of flight (t). The initial vertical speed is zero, and the vertical distance (y) traveled is 70 m. The acceleration is that of gravity.

y = 1/2 . a. t²

t = √(2y/a) = √(2 . 70 m/ 9.8 m/s²) = 3.8 s

The horizontal movement is a uniform motion (constant speed). The  horizontal speed is that of the plane. The horizontal distance at what the pilot should drop the weight is:

d = v . t = 44 m/s . 3.8s = 1.7 × 10² m

Final answer:

The mass of the ice block that requires a heat transfer of 9.8 x 10^5 J to convert it from -12°C ice to 12°C water is approximately 2.94 kg.

Explanation:

The question relates to the heat transfer required to convert a block of ice at -12°C to water at 12°C. To calculate the mass of the ice, we can use the formula for heat transfer — Q = mLf, where Q is the heat in joules, m is the mass in kilograms, and Lf is the latent heat of fusion for ice, which is 334 kJ/kg (or 334 x 10³ J/kg).

To find the mass of the ice, we set the total heat transfer equal to Q = mLf. The total heat transfer given is 9.8 × 10µ J. Using the latent heat of fusion value, we have:

9.8 × 10µ J = m × 334 × 10³ J/kg

Solving for m, we get:

m = µ ≟ 334 × 10³ kg

m ≈ 2.94 kg.

Answer:chemical weathering

Explanation:I just got a 100 percent on my quiz

Answer;

Diffraction

Explanation;

Diffraction involves a change in direction of waves as they pass through an opening or around a barrier in their path. Water waves have the ability to travel around corners, around obstacles and through openings. This ability is most obvious for water waves with longer wavelengths.

Diffraction of water waves is observed in a harbor as waves bend around small boats and are found to disturb the water behind them. The same waves however are unable to diffract around larger boats since their wavelength is smaller than the boat.

There are 4 ways in which electrons are emitted from the conductor.

i. Thermionic emission

ii. Electric field electron emission

iii. Photoelectric emission

iv. Secondary emission

In thermionic emission large amount of external energy in the form of heat is supplied to release free electrons from the metal.

In electric field electron emission, electrons are emitted from the metal surface when the metals are placed in a very strong electric field.

During photoelectric emission, light is absorbed by the metals and this provides energy to the valence electrons which break their bond with the parent atom and which are then released from the atom.

Valence electrons do have some kinetic energy, but they don't have enough energy to escape from the atom. During secondary emission, a high-speed electron is bombarded with an atom, which provides the energy for the valence electrons to break their bonds with their parent atom which are then released from the atom.

Answer:

Conditions that result in the emission of electrons from a conductor:

Heating the conductor to a suitable temperature

Exposing the conductor to a strong light

Subjecting the conductor to a very high applied voltage

Subjecting the conductor to high-speed electrons from another source

Explanation:

Given:

u= 0.91 m/s

a=0.31m/s∧2

s= 0.61 m

s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

a is the acceleration

Substituting the values

0.61=0.91×t +(0.31 ×t∧2)/2

0.61=0.91 t + 0.155 t∧2

t=0.61 secs

Consider the equation

v=u +at

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

Substituting the values we get

v= 0.91 +(0.31×0.61)

v= 1.099 m/s

Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

g is the acceleration due to gravity

h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.

When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy

[tex]U = mgh[/tex]

so when height of the car becomes half then its potential energy is given as

[tex]U_f = \frac{mgh}{2}[/tex]

so final potential energy when car falls down by half of the height will become half of the initial potential energy

So it is U = 50 MJ after falling down

Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car

So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ

so we can say at this point magnitude of potential energy and kinetic energy will be same

A. the same as the potential energy at that point.

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

[tex]\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}[/tex]

so that its period is [tex]T=3.644\,\frac{\mathrm s}{\mathrm{rev}}[/tex] (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration [tex]\Delta t[/tex]. Denote by [tex]\theta[/tex] the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

[tex]\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ[/tex]

[tex]\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ[/tex]

[tex]\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ[/tex]

We can then compute the magnitude of the velocity vector differences [tex]\Delta\vec v[/tex] for each time interval by using the law of cosines:

[tex]|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta[/tex]

[tex]\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]

and in turn we find the magnitude of the average acceleration vectors to be

[tex]\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

The average acceleration, a⃗ av, of the poodle running in a circle can be calculated using the formula a⃗ av = v⃗ ^2 / r. The direction of the acceleration is always towards the center of the circle in uniform circular motion. The calculations remain the same regardless of the time interval Δt.

This question is about the physics of circular motion, specifically calculating the acceleration of an object moving in a circular path. The poodle running in a circle at a constant speed indicates that this is a case of uniform circular motion. In such a situation the acceleration, a⃗ av, is always directed towards the center of the circle. This type of acceleration is also known as centripetal acceleration.

The formula to calculate the magnitude of the average acceleration in circular motion is a⃗ av = v⃗ ^2 / r, where v⃗  is the velocity (given as 5.00 m/s) and r is the radius (given as 2.9 m).

A) For Δt = 0.4 s, a⃗ av = (5.00 m/s) ^2 / 2.9 m = 8.62069 m/s^2, to four significant figures.

B) In uniform circular motion, the direction of the acceleration is always towards the center of the circle which is perpendicular to v⃗ 1.

C) & D) The calculations are identical for any Δt in uniform circular motion. so a⃗ av remains same and direction is also same.

E) & F) Again a⃗ av = 8.62069 m/s^2, to four significant figures and the direction is towards the center of the circle.

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3 a )Given:

u( initial velocity):60Km/hr=16.67m/sec

v(final velocity):120Km/hr=33.33m/sec

a(acceleration):20 m/s^2

Consider s as the distance traveled by the car. We can calculate s from the below formula.

v^2 - u^2= 2as

Where v is the final velocity measured in m/s

u is the initial velocity measured in m/s

a is the acceleration measured in m/s^2

s is the distance traveled by the car.

Substituting the given values in the above formula we get

33.33^2- 16.67^2= 2 x 20 x s

832.99= 40 s

s = 250.12 m

3b) Consider t as the time taken for the car to travel the above distance. We can calculate t from the below formula.

s = ut +1/2(at^2)

250.12= 16.67 X t + 1/2(20 x t^2)

500.233 = 33.33t + 20t^2

Solving the above quadratic equation we get t= 1026 secs.

4) Given

v( final velocity) = 0.

time taken to cover the distance= 25 secs

Distance traveled(s)=40Km= 40000m

Now consider the below equation

v = u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Substituting the given values in the above equation we get

0= u+ax25

u= -25a

Now we already know that

s = ut + 1/2(at^2)

Where s is the distance traveled

u is the initial velocity

a is the acceleration

t is the time

Substituting the given values in the above formula we get

40000 = u25 +1/2(ax25x25)

Now as solved above -25a =u. Substituting this in the above formula we get

40000= 25u +1/2(-25u)

40000= 12.5u

Thus u = 40000/12.5

u = 3200m/s

As per the above derived equation

We know in this case

-25 a = u

a= -128 m/s^2

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

F= ma

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

v=u +at

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs

s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

s= 200 m

Answer:

Stopping distance, s = 200 meters

Explanation:

Mass of the car, m = 2000 kg

Force acting in the car, F = -2000 N

Initial speed of car, u = 20 m/s

Finally, it stops, v = 0          

Using second equation of motion as :      

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{-2000}{2000}[/tex]

[tex]a=-1\ m/s^2[/tex]

Let s is the stopping distance. Now using third equation of motion as :

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0-(20)^2}{2\times -1}[/tex]

s = 200 meters

So, the stopping distance of the car is 200 meters. Hence, this is the required solution.

The lysosomes functions like a recycling center. The pH inside the lysosome is acidic around 5 due to the hydrogen ions and protons produced inside. These lysosomes contain hydrolase which is an enzyme which breaks down the molecular bonds in a chemical compound.This enzyme is active only in acidic conditions.

the answer is mitochondria

Quality Factor is defined as the ratio of Resonance frequency per unit band width

it is given as

[tex]Q = \frac{\omega_0}{\Delta \omega}[/tex]

[tex]Q = \frac{2\pi f_0}{2\pi f_{fw}}[/tex]

now here it is given that

[tex]f_0 = 1.40 hz[/tex]

[tex]f_{fw} = 0.021 hz[/tex]

now we have

[tex]Q = \frac{2\pi*1.40}{2\pi*0.021}[/tex]

so we have

[tex]Q = 66.67[/tex]

So quality factor of given AC is 66.67

Cellular respiration is of two types namely aerobic and anaerobic respiration.

During aerobic respiration,the food(glucose) is broken down in the mitochondria to produce ATP,water and CO2.This is carried out in 3 steps glycolysis followed by Tricarboxylic acid cycle and electron transport chain.

During anaerobic respiration, glucose is broken down in the absence of oxygen to produce ethanol,CO2 and water. Anaerobic respiration occurs in the cytoplasm.

Answer:

Salt Compound

Explanation:

acid + base  →→→ water +salt compound

Given:

m(mass of the car)=1134 Kg

u(Initial velocity)=83Km/HR=23m/s

s(distance traveled by the car)=98m

v(final velocity)=0(as it is given the car stops).

Now we know,

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

0=23+at

at=-23

Also

s=ut+1/2(at^2)

s is the distance covered by the car

u is the initial velocity

t is the time necessary for the car to cover a particular distance.

a is the acceleration

Now substituting these values we get

98=23t-1/2(23t)

98=23t-11.5t

11.5t=98

t=8.52secs

Now we have already derived

at=-23

ax8.52=-23

a=-23/8.52

a=-2.75 m/s^2

F=mxa

Where F is the force acting on the car.

m is the mass of the car.

a is the acceleration.

F=1134 x-2.75

F=-3119N

A) They all orbit the sun

because they all are attracted through the forces of gravity.

Formula of the gravitational force between two particles:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

[tex]G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}[/tex] is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

[tex]F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N[/tex] to the right

The gravitational force exerted by particle C on particle A is

[tex]F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N[/tex] to the right

So the net gravitational force on particle A is

[tex]F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N[/tex] to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

[tex]F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N[/tex] to the left

The gravitational force exerted by particle C on particle B is

[tex]F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N[/tex] to the right

So the net gravitational force on particle B is

[tex]F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N[/tex] to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

[tex]F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N[/tex] to the left

The gravitational force exerted by particle B on particle C is

[tex]F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N[/tex] to the left

So the net gravitational force on particle C is

[tex]F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N[/tex] to the left

Particle A,B, and C is  [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively. Negative sign represents left direction.

The gravitational force between two bodies

[tex]\rm \bold { F= G\frac{m^1 m^2}{r^2} }[/tex]

Where,

G- gravitational constant = [tex]\rm \bold{6.67\times 10^1^1 Nm^2kg^-^2 }[/tex]

[tex]\rm \bold { {m^1 m^2} }[/tex] = mass of bodies

r - is distance between them

Net gravitational force on Particle A

[tex]\rm \bold{F_a = F_b+ F_c}[/tex]

The gravitational force exerted by particle B on particle A is [tex]\rm \bold{ 5.14 \times 10^-^5N}[/tex] to the right .

The gravitational force exerted by particle C on particle A is [tex]\rm \bold{ 5.6 \times 10^-^6N}[/tex] to the right.

Hence, net Gravitational force on A is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex]

Net gravitational force on Particle B

[tex]\rm \bold{F_b = F_a+ F_c}[/tex]

The gravitational force exerted by particle A on particle B is [tex]\rm \bold{ -5.14 \times 10^-^5N}[/tex] on to the left.

The gravitational force exerted by particle C on particle B is [tex]\rm \bold{ 8.41 \times 10^-^5}[/tex] to the right.

Hence net  gravitational force on particle B will be [tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex]

Net gravitational force on Particle C is

[tex]\rm \bold{F_c = F_a+ F_b}[/tex]

The gravitational force exerted by particle A on particle C is [tex]\rm \bold{ -5.6 \times 10^-^6N}[/tex] to the left.

The gravitational force exerted by particle B on particle C is [tex]\rm \bold{ -8.41 \times 10^-^5N}[/tex] to the left.

Hence,  net gravitational force on particle C is [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] to the left.

Therefore we can conclude that particle A,B, and C is  [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively.

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Quite a few things can be determined using this information.

(i) First, we can calculate the Weight of the fireman using W = mg.

We get its magnitude as W = (80)(9.8) = 784 N

(ii) While his Weight is responsible for pulling him down, the pole exerts a constant Resistive Force that is given to be 500 N.

We can calculate the Net Force acting on the fireman as

[tex]F_{net}  = W - Resistive Force[/tex]

We get its numerical value to be [tex]F_{net}  = 284N[/tex]

(iii) Using this, we can calculate the acceleration with which he slides down the pole from Newton's 2nd law equation as

[tex]F_{net} =ma[/tex]

Therefore, [tex]a = \frac{284}{80} =3.55 m/s^{2}[/tex]

(iv) With this acceleration, he slides down a distance of 5.0 m - 0.4 m = 4.6 m before he starts applying an additional force with his knees.

We can calculate the velocity he attains just before bending his knees using the following data:

Initial Velocity at the top of the pole [tex]V_{i} =0[/tex]

Vertical displacement down the pole D = 4.6 m

Acceleration [tex]a = 3.55 m/s^{2}[/tex]

Final Velocity [tex]V_{f}  = ?[/tex]

Using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex]

Plugging in the numbers, we have [tex]V^{2} _{f} =0+2(3.55)(4.6)[/tex]

Thus, we get the value of [tex]V_{f}[/tex] as 5.72 m/s

(v) This velocity serves as the initial velocity for the part of the journey with his knees bent.

We can calculate the acceleration he has using the following data:

Initial Velocity [tex]V_{i}=5.72[/tex] m/s

Final Velocity [tex]V_{f} =0[/tex]

Displacement during the last part of the journey D = 0.4 m

Acceleration a = ?

Again using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex], and plugging in the known numbers, we get

[tex](0)^{2} =(5.72)^{2} +2(a)(0.4)[/tex]

Hence, [tex]a=-40.898 m/s^{2}[/tex]

(vi) We can calculate the Net Force acting on him during this part of the journey as

[tex]F_{net} =(80)(-40.898)=-3271.84N[/tex]

(vii) Since the Net Force is the vector sum of Weight, Resistive Force, and the additional Knee Force, we can write them as

[tex]W-Resistive Force-Knee Force=-3271.84N[/tex]

Solving for Knee Force gives its magnitude to be 3555.84 N.

Thus, the fireman begins applying an additional 3,556 N force to stop himself in 0.4 m.

(viii) We can even calculate the time taken for the entire journey. We will deal with it in two parts.

For part one, the following information can be used:

Initial Velocity [tex]V_{i} =0[/tex]

Final Velocity [tex]V_{f} =5.72m/s[/tex]

Acceleration [tex]a = 3.55m/s^{2}[/tex]

Time taken t = ?

Using the equation [tex]V_{f} =V_{i} +at[/tex], we get the time taken as 1.6 seconds.

For the second part of the journey, the following information can be used:

Initial Velocity [tex]V_{i} =5.72m/s[/tex]

Final Velocity [tex]V_{f} =0[/tex]

Acceleration [tex]a=-40.898m/s^{2}[/tex]

Time taken t = ?

Using the equation [tex]V_{f} =V_{i} +at[/tex] again, and plugging in the appropriate values, we get the time taken as 0.14 seconds.

Hence, the total time the fireman took to slide down the 5.0 m pole is 1.74 seconds.

Hope this helps!

The calculation using the physics concept of work and energy determines the work done by a fireman when he slides down and stops, and quantifies the force he exerts when slowing to a stop.

This is a classic problem in physics, specifically in the area of kinetic energy and work-energy theorem. First, as the fireman slides down, he is doing work and transforming potential energy into kinetic energy. We can calculate this work using the equation work = force x distance. In this case, the force is the resistive force of 500N, and the distance is 5.0m. So the work done is 500N x 5.0m = 2500J.

Next, when he bends his knees to slow to a stop, he is doing another amount of work to remove this kinetic energy. We can use the same formula as before, but this time the distance is 0.40m. We won't know the force directly, but we know the work done needs to be equal to the kinetic energy obtained earlier. So, we can solve for the 'force': 2500J = force x 0.40m, thus the force exerted would be around 6250N.

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Venus is your best bet looking at your answers.

Answer:

Venus

Explanation:

Solar system has 8 planets namely Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. Out of these the first four planets are rocky while the last four are gaseous. When you compare the size of the rocky planet with gaseous planets, gaseous planets are way too bigger than rocky planets.

Using the same information, out of the given options Venus is the correct answer as it is smaller than other three planets and has a rocky surface and atmosphere.

Storage form of energy:

Potential energy:

All stationary objects are having potential energy stored in it. This energy can be transferred in form of kinetic energy when it comes in the motion from rest. Example, An object placed at height h having potential energy in it. When it comes in motion from the rest the potential energy is converted into kinetic energy.

Nuclear energy:

Nuclear energy is energy that is stored in nucleus of any element. Example, fusion reaction on sun gives earth solar energy.

Electric energy:

Electrical energy is due to movement of the electrical charges. Example, In elctrical batteries electrical energy is stored.

Thermal energy:

Thermal energy is the internal energy of a substance that is transferred to other substance in the form of heat. Example, on heating water is a beaker stem energy is developed.

Magnetic energy:

Magnetic energy is the potential energy stored in the magnetic field. Example, using magnetic energy electric field is produced according to Faraday's law.

Energy has various stored forms, including chemical, mechanical, radiant, electrical, and gravitational potential energy.

Energy can be categorized into multiple storage forms. We often see energy transforming from one form to another in everyday situations.

Chemical Energy: This is stored in the bonds between atoms and molecules. For instance, the energy stored in the bonds of a glucose molecule that our bodies break down for fuel.

Mechanical Energy: It's the sum of kinetic and potential energy associated with the motion and position of an object. A working windmill is an example of mechanical energy.

Radiant Energy: It's the energy of electromagnetic waves. For example, the light emitted by the sun.

Electrical Energy: This involves the movement of charged particles. An example would be the energy that drives your computer or mobile device.

Gravitational potential energy: It's depends on an object's height above the ground. The potential energy of a book on a shelf, for example, becomes kinetic energy when the book falls, which proves the energy's existence.

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If a boy is continuously accelerated then we can assume it an example of uniform acceleration

Here we can say

[tex]d = v_i*t + \frac{1}{2}at^2[/tex]

now here we have

[tex]d = v*t + 0.5a t^2[/tex]

if we draw the graph between d and t then we can say that this graph will be a quadratic graph as it is the equation with power 2

So here in this case the graph will be QUADRATIC

Option "D" is correct option

If the interaction is gravitational or electrical, it gets multiplied by (1/2-squared) or 1/4 .

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This is mathematically represented as

F= (G X m1 x m2) /r∧2

where F is the force acting between the charged particles

r is the distance between the two charges measured in m

G is the gravitational constant which has a value of 6.674×10^-11 Nm^2 kg^-2

m1 and m2 are the masses of the objects measured in Kg

Now if the distance between the is doubled then r becomes 2r. Substituting this in the above formula we get the new Force as

Force (new) = (G X m1 x m2) /(2r)∧2

Thus dividing Force(new)/Force we get

Force(new)/Force = 1/4.

Thus the gravitational force becomes 1/4th of the original value if the distance between the two masses are doubled.