All categories

3 years ago

A 100 kg individual consumes 1200 kcal of food energy a day. Calculate

Physics

1 answer:

IrinaK [193]3 years ago

4 0

Answer:

(a) 5142.86 m

(b) 317.5 m/s

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 = 1/2 x 100 x v^2

v = 317.5 m/s

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

You might be interested in

Which one of these is a chemical property of a substance?

Schach [20]

Your answer is D Flammability

6 0

3 years ago

What is the force on a 1000 kg elevator that is free falling at 9.8 m/s/s?

Anton [14]

Answer:

Explanation:

Force = (mass) · (acceleration)

= (1,000 kg) · (9.8 m/s²)

= 9,800 newtons

4 0

3 years ago

According to Freud, a person’s current perceptions and thoughts would be located at the __________ level. A. conscious B. noncon

Nikolay [14]

Answer:

A. conscious

Explanation:

8 0

3 years ago

The earth rotates once per day about its axis. where would you have to stand in order to have the smallest tangential speed?

Yuliya22 [10]

<span>The earth as a whole rotates with a fixed angular velocity. Therefore, you need to be at the rotation axis to have the smallest tangential speed. This would be at the north or south pole.</span>

5 0

3 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

4 years ago