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Sophie [7]
3 years ago
5

Write the law of conservation of energy. How does it apply to eating and exercising?

Physics
1 answer:
zzz [600]3 years ago
5 0
The Law of Conservation of energy states that the total energy of an isolated system remains constant-it is said to be the conserved over time. It applies by us burning calories because humans can lose or gain weight.
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9- Un auto <br> Recorre 500m en 4 segundos, calcula la velocidad promedio.
Brut [27]

Answer:

Is this Spanish?

Explanation:

7 0
3 years ago
While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds
Ksju [112]

Answer:

  P₂ = 2 P₁

we conclude that in the second time the power used is double that in the first rise

Explanation:

In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.

Now we can analyze the required power,

         P = W / t

From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,

therefore the first time the power is

           P₁ = E / 10

           P₁ = 0.1 E

for the second time the power is

          P₂ = E / 5

          P₂ = 0.2 E

we see that the power in the second case is

         P₂ = 2 P₁

Therefore, we conclude that in the second time the power used is double that in the first rise.

6 0
3 years ago
A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
Gekata [30.6K]
(a) The stone moves by uniform accelerated motion, with constant acceleration g=9.81 m/s^2 directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
y(t)= y_0 -  \frac{1}{2}gt^2= 750 - 4.9 t^2

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
750-4.9 t^2 = 0
from which we find the time t after which the stone reaches the ground:
t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s

(c) The velocity of the stone at time t can be written as
v(t) = -gt
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

(d) if the stone has an initial velocity of v_0 = 6 m/s, then its law of motion would be
y(t)=y_0 - v_0t -  \frac{1}{2}gt^2
and we can find the time it needs to reach the ground by requiring again y(t)=0:
0=750 - 6t - 4.9 t^2
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0
3 years ago
A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog
Hunter-Best [27]

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

6 0
3 years ago
8. A man takes one minute to lift 4 bags of sugar each of weight 50N
timofeeve [1]

Answer:

D

Explanation:

the formula is F=mgh

so now you can write it like

m= 4×50=200

200×1.5×10=30000

6 0
3 years ago
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