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3 years ago

Proof of conservation of energy

1 answer:

6 0

If you compare the energy at two different times in an equation, you'll see no difference, because it's been conserved. The total energy of a system is the sum of its energy in motion - its kinetic energy - and its energy due to its position, which is its potential energy

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A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step

Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

7 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago

A stone is thrown vertically upward with velocity 9.8ms 1)calculate the time taken to reach maximum height.2)what is the maximum

ira [324]

Answer:

1.t=-1.96sec

2.H=4.8m

3.T=1.96sec

4.R=19.2m

Explanation:

u=9.8,t=?,sin theta=1

using formula t=2usintheta/g

t=2x9.8x1=19.6/10

t=1.96seconds

using formula H=u(squared)sin(squared)theta/2g

H=9.8(squared)x1(squared)/2x10

H=96x1/20

H=96/20

H=4.8m

using formula T=2usintheta/g

T=2x9.8x1/10

T=19.6/10

T=1.96sec

using the formula R=u(squared)sin2theta/g

R=9.8(squared)x2/10

R=96x2/10

R=192.08/10

R=19.2

5 0

3 years ago

Read 2 more answers

a small negatively charged sphere with a mass of 5.4*10^-5 is suspended between two parallel plates. the potential difference is

labwork [276]

Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s² [ constant value for earth system ]

Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶ [ negative sign represents the nature of charge ]

So, Your Final answer would be 1.47 × 10⁻⁶

Hope this helps!

3 0

3 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$