Based on what you learn about gravity describe why objects float in outer space

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   [tex]COP  =  \frac{T_i}{T_o - T_i}[/tex]

Here [tex]T_i[/tex] is the inside temperature

while  [tex]T_o[/tex] is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         [tex]Q \ \alpha \ (T_o - T_i)[/tex]

=>        [tex]Q= k (T_o - T_i)[/tex]

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   [tex] \frac{T_i}{T_o - T_i}[/tex]  amount of heat

   E  unit of electricity will remove  [tex]Q= k (T_o - T_i)[/tex]

So

      [tex]E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]

=>   [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]

given that  [tex]\frac{k}{T_i}[/tex] is constant

    =>  [tex]E \  \alpha  \  (T_o - T_i)^2[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  [tex]T_i   =  30 ^oC[/tex]

 and      [tex]T_o  =  40 ^oC[/tex]

Hence  

     [tex]E = K (T_o - T_i)^2[/tex]

Here K stand for a constant

So  

        [tex]E = K (40 -  30)^2[/tex]

=>      [tex]E = 100K [/tex]

Now if  the  [tex]T_i   =  20 ^oC[/tex]

Then

       [tex]E = K (40 -  20)^2[/tex]

=>      [tex]E = 400 \ K[/tex]

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       [tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]

So

       [tex]E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]

=>   [tex]E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]

Assuming [tex]\frac{k}{T_i}[/tex] is a constant

Then  

     [tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

[tex]v=gt[/tex]

And the distance traveled downwards is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]

Replacing into the first equation:

[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]

Rationalizing:

[tex]\displaystyle v=\sqrt{2gy}[/tex]

Let's call v1 the final speed of the package dropped from a height H. Thus:

[tex]\displaystyle v_1=\sqrt{2gH}[/tex]

Let v2 be the final speed of the package dropped from a height 4H. Thus:

[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]

Taking out the square root of 4:

[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]

Dividing v2/v1 we can compare the final speeds:

[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]

Simplifying:

[tex]\displaystyle v_2/v_1=2[/tex]

The final speed of the second package is twice as much as the final speed of the first package.

Answer:

54mph

Explanation:

Answer:

It says the answer. (5 marks.)

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

Answer:

The answer is

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

From the question

velocity = 1.45 × 10² m/s

time = 1.8 × 10³ s

We have

distance = 1.45 × 10² × 1.8 × 10³

We have the final answer as

[tex]2.61 \times {10}^{5} \: \: m[/tex]

Hope this helps you

Answer:

Explanation:

The acceleration of an object given it's velocity and time taken acting on it can be found by using the formula

[tex]acceleration = \frac{velocity}{time} \\ [/tex]

From the question

time = 4 s

velocity = 10 m/s

So we have

[tex]acceleration = \frac{10}{4} = \frac{5}{2} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

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Explanation:

Answer:

The distance will be x = 1320 [m]

Explanation:

To solve this problem we must use the expression of physics that relates space to time, which is defined as speed.

v = Speed = 11 [m/s]

t = time = 120 [s]

x = distance [m]

v = x/t

x = v*t

x = 11*120

x = 1320 [m]

Answer:

We are given:

mass of the truck (m) = 3000 kg

velocity of the truck (v) = 25 m/s

Calculating the Momentum:

We know that the momentum of an object is equal to its mass times it's velocity

P = mv           (where P is the momentum of the truck)

Momentum of the truck:

replacing with the values of the truck

P = mv

P = 3000 * 25

P = 75000 kg m/s

Therefore, the momentum of the truck is 75000 kg m/s

The acceleration/deceleration of the truck is 11.56 m/s^2

Velocity = change in displacement/ time

Acceleration = change in velocity/ time.

The acceleration, velocity and distance travelled of a body is related by the equation below:

where v is the initial velocity

u is the final velocity

a is the acceleration, and

s is the distance travelled

From the data provided:

v = 0

u = 34 m/s

s = 50 m

a = ?

making a subject of the formula

a = -u^2/2s

a = - 34^2 / 2 × 50

a = -11.56 m/s^2

Therefore, the acceleration/deceleration of the truck is 11.56 m/s^2

Learn more about velocity and acceleration at: https://brainly.com/question/742413

Explanation:

When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.

Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.

Answer:

a) 4.0 s

b) 16 m/s

c) The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s

d) Equal distance are covered between 0 - 4 s and 4 -5 s

Note: The question is incomplete. The complete question is as follows;

Refer to the chart below that has data about a moving object to answer the following questions.

Time Elapsed         0.0s   1.0s    2.0s    3.0s    4.0s    5.0 s

Distance Traveled 0.0m 10.0m 20.0m 30.0m 40.0m 80.0 m

a. What is the elapsed time between the 0-m mark and the 40-m mark?  

b. How large is the average velocity of the object for the interval from 0-5 s ?  

c. How does the interval of 3-4 s compare with the interval from 4-5 s in terms of distance?    

d. How does the interval of 0-4 s compare with the interval from 4-5 s in terms of distance?    

Explanation:

a. From the data provide, time elapsed between the 0 m - 40 m mark is 4.0 s

b. Average velocity = total distance/ total time

average velocity = 80 m/ 5.0 s = 16.0 m/s

c. Distance covered between 3 - 4 s = 40 m - 30 m = 10 m

Distance covered between 4 - 5 s = 80 m - 40 m = 40 m

The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s

d. Distance covered between 0 - 4 s = 40 m - 0 m = 40 m

Distance covered between 4 - 5 s = 80 m - 40 m = 40 m

Equal distance are covered between 0 - 4 s and 4 -5 s

the measure of how many a wave passes on a certain point in a certain time is

frequency

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² -  2gh

0 = u² -  2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

Complete Question

A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?

[tex]A.\ \ P_2=4P_1[/tex]

[tex]B.\ \  P_2=\frac{4}{3} P1[/tex]

[tex]C.\ \  P_2=P_1[/tex]

[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]

[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]

Answer:

The correct option is  B

Explanation:

From the question we are told that  

    The mass of the brick is  M

    The  height height of the 10th floor is  H =  4y

     The height attained during the first part of the lift is  [tex]h_1 =  3y[/tex]

     The time taken is [tex]t_1 =  4T[/tex]

    The height attained during the second part of the lift is  [tex]h_2  = y[/tex]

    The time taken is  [tex]t_2  =  T[/tex]

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_1}{t_1}[/tex]

=>      [tex]v_1  =  \frac{3y}{4T}[/tex]

Generally the velocity of the crane during the first lift is mathematically represented as

         [tex]v_1  =  \frac{h_2}{t_2}[/tex]

=>      [tex]v_1  =  \frac{y}{T}[/tex]

Generally the power generated during the first lift is  

     [tex]P_1 =  F_1 *  v_1[/tex]

Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as

       [tex]F_1  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_1 =  Mg * \frac{3y}{4T}[/tex]

Generally the power generated during the second lift is  

     [tex]P_2 =  F_2 *  v_2[/tex]

Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as

       [tex]F_2  =  M  *  g[/tex] here g is the acceleration  due to gravity

So

       [tex]P_2 =  Mg * \frac{y}{T}[/tex]

So the ratio  of the first power to the second power is  

      [tex]\frac{P_1}{P_2}  =  \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]

=>   [tex]\frac{P_1}{P_2}  = \frac{3}{4}[/tex]

=>    [tex]P_2 = \frac{4}{3} P_1[/tex]

Answer:engineers developed and improved solutions in response to consumer demand for equal access for people with disabilities.

Explanation:

Apex

Answer:

The displacement of a body has two components: rigid-body displacement and deformation. A rigid-body displacement consists of a simultaneous translation and rotation of the body without changing its shape or size.

Hope this helps :)

Answer:

1, 1583.33 V/m

2, 4.72*10^-12 C

3, 39.2*10^-11 C

Explanation:

1

E = V / d

E = 8.55 / 5.4*10^-3

E = 8.55 / 0.0054

E = 1583.33 V/m

2

Capacitance, C = (k * e0 * A) / d, where k = 1

A = area of capacitor, 3.37 cm² = 3.37*10^-4 m²

d = plate separation, 5.4 mm

e0 = Constant, 8.85*10^-12

Applying these, we have

C = (1 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3

C = 29.82*10^-16 / 0.0054

C = 5.52*10^-13 F

Since Q = CV, then

Q = 5.52*10^-13 * 8.55

Q = 4.72*10^-12 C

3

We are given that k = 83, so

Capacitance, C = (k * e0 * A) / d

C = (83 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3

C = 2.475*10^-13 / 0.0054

C = 4.58*10^-11 F

Q = CV

Q = 4.58*10^-11 * 8.55

Q = 39.2*10^-11 C

Answer:

60 kg

80 kg

Explanation:

Work is equal to the change in energy.

W = ΔE = E − E₀

Let's start with block B.  The work done by the tension force is equal to the change in energy.  Initially, the block has potential energy.  Finally, the block has kinetic energy.

W = ΔE

FΔy = ½ mv² − mgh

T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)

T (-2.0 m) = m (-2 m²/s²)

T = m (1 m/s²)

Now let's look at block A.  The work done by tension and against friction is equal to the change in energy.  Initially, the block has no energy.  Finally, it has both kinetic and potential energy.

W = ΔE

Fd = ½ mv² + mgh − 0

(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)

(T − Nμ) (2.0 m) = 120 J

T − Nμ = 60 N

Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.

N = mg cos θ

N = (4.00 kg) (10 m/s²) (⅘)

N = 32 N

Substitute:

T − 32μ = 60 N

If μ = 0, then T = 60 N and m = 60 kg.

If μ = ⅝, then T = 80 N and m = 80 kg.

Answer:did you find you answer also i answered so you can give him brainiest

Explanation:

Answer: the link isnt loading

Answer:

it dies

Explanation:

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Answer:

D. Conclusion.

Explanation:

Answer:

Explanation:

if you bounce a tennis ball, it will come back up as an equal reaction to your action (dropping it). there is an equal but opposite reaction in this scenario, demonstrating newton's third law.

Answer:

when u jump in the air and hit the ground the ground applies the same force back which applies a opposite force which causes u to propel into the air

Explanation:

hope this helped :)

Answer:

1.84m

Explanation:

Given parameters:

Initial velocity  = 6m/s

Unknown:

height of jump  = ?

Solution:

To solve this problem, we have to apply the right motion equation:

        V²  = U²  - 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity  = 9.8m/s²

H is the height

   Final velocity is 0

  Solve;

           0² = 6² - 2x9.8xH

           -36  = -19.6H

               H = 1.84m

Answer:

D) Lifting it to a high shelf

Explanation:

Potential energy is where energy could be released at any moment. For example, if a cat gets on a tree it's creating potential energy, once that cat jumps from the tree the energy releases. D is the answers because if you set the book in a high shelf, you are only creating more potential energy.

hope I explain my self well enough

cheers man and good luck

Answer:

The value is [tex]E_i = 1.5596 *10^{-18} \ J[/tex]

Explanation:

From the question we are told that

The wavelength is [tex]\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m[/tex]

The velocity is [tex]v = 2.371*10^6 \ m/s[/tex]

The mass of electron is [tex]m_e = 9.109*10^{-31} \ kg[/tex]

Generally the energy of the incident light is mathematically represented as

[tex]E = \frac{h * c}{\lambda}[/tex]

Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

h is the Planck constant with value [tex]h = 6.62607015 * 10^{-34 } J\cdot s[/tex]

So

[tex]E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}[/tex]

=> [tex]E = 4.12 *10^{-18} \ J [/tex]

Generally the kinetic energy is mathematically represented as

[tex]E_k = \frac{1}{2} * m_e * v^2[/tex]

=> [tex]E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2[/tex]

=> [tex]E_k = 2.56 *0^{-18} \ J [/tex]

Generally the ionization energy is mathematically represented as

[tex]E_i = 4.12 *10^{-18} - 2.56 *0^{-18}[/tex]

=>     [tex]E_i  =  1.5596 *10^{-18} \  J[/tex]