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nydimaria [60]
1 year ago
6

Which graph below represents how the velocity of the sphere changes over time when falling with constant acceleration?

Physics
1 answer:
Amiraneli [1.4K]1 year ago
4 0

Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.

  • Acceleration is the measure of how quickly a body's velocity varies with regard to time, and constant acceleration occurs when a body's velocity changes proportionately over a period of time, or at a constant rate. It measures in m/s2.
  • It is claimed that a body has continual positive acceleration when it begins to move with an initial velocity of zero and gradually increases to a positive value over time.
  • Constant positive acceleration is demonstrated by a ball falling freely in a vertical direction.

To know more about  constant acceleration. visit : brainly.com/question/9754169

#SPJ1

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At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assum
Verdich [7]

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

W_c=Mgh        (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J     (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

0.16(E_c)=4,164,294.4J

Ec: Calories

You solve for Ec and convert the result to Cal:

E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal

The amount of Calories needed by the climber was 6220.56 kcal

4 0
3 years ago
How do I work out the percentage??
NeX [460]
Percentage for which columns btw if its something as easy as percentage for 4/5... you divid 4/5= .80 then.... .80*100= 80%
7 0
3 years ago
Read 2 more answers
SHOW ADEQUATE WORKINGS IN THIS SECTION
cluponka [151]

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

3 0
3 years ago
____ charges are attract each other
timama [110]

Answer:

One positive and one negative

3 0
3 years ago
Read 2 more answers
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

6 0
3 years ago
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