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egoroff_w [7]
3 years ago
6

Jackie rides her bike to school every day and enjoys playing beach volleyball on weekends. She also helps with tasks around the

house, like pulling weeds in the garden and sweeping the porch. Which of the following describes a possible benefit of Jackie's physical activity? (1 point) a Jackie may be less likely to commit to a healthy future. b Jackie may have an increased chance of disease. c Jackie may maintain healthy body fat levels. d Jackie may require more assistance performing daily tasks.
Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

The correct answer is - C) Jackie may maintain healthy body fat levels.

Explanation:

Jackie has a very active daily and on weekends routine that includes riding the bike, playing volleyball, pushing weeds, and sweeping the porch. The activities that Jackie is performing daily as well as the time that she spends on exercise may help her to maintain body fat levels.

Being physically active helps in maintaining fat levels in the body of an individual, and it is healthy for the individual.

Thus, the correct answer is - C) Jackie may maintain healthy body fat levels.

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A conductor of radius r, length l and resistivity p has resistance R. It is melted down and formed into a new conductor, also cy
Vaselesa [24]

Answer:

b) R/4 (There seems to an error in mentioning the multiple choices of this question, please see below explanation of correct calculations for this  question.)

Explanation:

dimension of the conductor before melting is l, r

reistivity is p

R=(p*l)/(pie*r2)

after reforming length is reduced to L=l/4

volume in both the cases will be same

i.e. pie * r^2 * l =pie * R^2 * L

r^2 * l = R^2 * (1/2)l

due to this radius will become R=sqrt(2) * r

now new reistance is given by Rx=(p * L)/(pie * R^2)

i.e. Rx=(p * l/2)/(pie * r^2 * 2)

after simplification RX=((p * l)/(pie * r^2))/4

i.e. Rx=R/4

5 0
3 years ago
Read 2 more answers
Using an analogy of a water balloon, describe the function of the bladder. (2 points)
a_sh-v [17]

Answer:

it fucntions

Explanation:

7 0
3 years ago
Read 2 more answers
You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The m
Alecsey [184]

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

                                   UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

                                      y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

                                      UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

                                      UG = UK

- Where UK is kinetic energy given by:

                                      UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

                                     m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

                                    Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

                                     Vx,f = sqrt (2*g*x*sin(Q))

8 0
3 years ago
Which of the following types of reactions would decrease the entropy within a cell?
kifflom [539]

Answer:

dehydration reactions

7 0
3 years ago
The cable of a crane is lifting a 750 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/s in a distance of 3.5 m
IgorC [24]

To solve this exercise it is necessary to apply the equations concerning Work, both by general definition and by conservation of energy.

In other words, the work done by an object due to gravity is the equivalent to that defined by the potential energy equations, that is

W= mg\Delta h

Where,

m=mass

g=gravitational acceleration

\Delta h = Change in height

On the other hand we have that the work done by tension is defined by the conservation of kinetic and potential energy, that is to say

W= \Delta KE + \Delta PE

Where,

\Delta KE = Change in Kinetic Energy

\Delta PE =Change in Potential Energy

PART A) As defined by the work done by gravity would be given by,

W = mg\Delta h

W = (750)(9.8)(3.5)

W = 25.725kJ

Therefore the work done by gravity is 25.725kJ

PART B) The work done by the tension applies the energy conservation equation, that is to say

W= \Delta KE + \Delta PE

W = (\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2)+(mgh_f-mgh_i)

W = \frac{1}{2}m(v_f^2-v_i^2)+mg(h_f-h_i)

Replacing with our values,

W = \frac{1}{2}(750)(0.75^2-0.25^2)+(750)(9.8)(3.5)

W = 25.912kJ

Therefore the work done by tension is 25.9kJ

5 0
3 years ago
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