answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vitek1552 [10]
2 years ago
14

A person ate 0.50 pound of cheese (an energy intake of 4000 kJ). Suppose that none of the energy was stored in his body. What ma

ss (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes 44.0 kJ to vaporize 1 mole of water.)
Physics
1 answer:
mamaluj [8]2 years ago
4 0

Answer:

The answer to the question is;

1637.769 grams of water will need to be perspired in order to maintain his original temperature.

Explanation:

Energy intake of the person = 4000 kJ

Energy required to vaporize 1 mole of water = 44.0 kJ

That is 44.0 kJ/mole

Therefore

The number of moles of water that can be vaporized by 4000 kJ is given by

(4000 kJ)/ (44.0 kJ/mole) = 90.91 moles.

Mass of one mole of water = Molar mass of water = 18.01528 g/mol

Since number of moles of water = (\frac{Mass .of. water}{Molar. mass. of. water})

We therefore have

Mass of water = (Number of moles of water)× (Molar mass of water)

Mass of water = 90.91 moles× 18.01528 g/mol = 1637.769 g

The mass (in grams) of water that he  would need to perspire in order to maintain his original temperature is 1637.769 g.

You might be interested in
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
patriot [66]

If I have done my math correctly, your answer is 67.5

4 0
2 years ago
Read 2 more answers
A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
2 years ago
Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a
EleoNora [17]
Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ...  V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1;  -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1;  -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)
3 0
2 years ago
Read 2 more answers
What is the magnitude of the relative angle φ
melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0
2 years ago
A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d
AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

5 0
1 year ago
Other questions:
  • If a neutral object such as paper comes close to a positively charged plastic rod, what type of charge accumulates on the side o
    14·2 answers
  • What is the total energy released when 9.11 x10^-31 ki?
    7·1 answer
  • A cat accelerates from rest to 10m/s when it sees a dog. This takes 2 seconds. What was the acceleration of the cat
    11·2 answers
  • An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
    10·1 answer
  • 3. A 64 lb weight stretches a spring 4 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given
    15·1 answer
  • A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalph
    13·1 answer
  • How much force is required to pull a spring 3.0 cm from
    7·1 answer
  • As Jackie pushes box props up the ramp to the stage, the frictional force will _________ her efforts. A) prevent B) speed up C)
    15·1 answer
  • A student starts at the origin and ends up at a position 500 meters north of the origin. She knows she walked 250 meters straigh
    11·1 answer
  • A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!