Burns produced by steam at 100°C are much more severe than those produced by the same mass of 100°C water. Calculate the quantity of heat in (Cal or kcal) that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C. Specific heat of water = 1.00 kcal/(kg · °C); heat of vaporization = 539 kcal/kg; specific heat of human flesh = 0.83 kcal/(kg · °C).

Answer:

Planks?

Explanation:

It's kinda resting

Answer:

weights

walking

stretching

etc.

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m  d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

a)

• Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
• However, linear speeds of points at different distances from  the center, are different.
• Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

• Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

• As we have already said, ωout = ωin = 3.7 rad/sec

b)

• Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
• Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
• 

        ⇒  Δθ₁ = Δθ₂ = 18.5 rad.

c)

• The linear distance traveled by each child, will be related with the linear speed of them.
• Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

      vout is a given of the problem ⇒ vout = 11. 5 m/s

• Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

d)

• The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

• Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

• In the same way, we get Fcout (the force on the boy near the outer edge):

e)

• The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
• The maximum friction force is given by the product of the coefficient of static friction times the normal force.
• Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
• As both boys have the same mass, the normal force is also equal.
• This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
• 

• If this force is greater than the centripetal force, the boy will be able to hold on.
• So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.

The given data in the problem is;

q is the charge= 1.6 ×10⁻¹⁹ C

V is the electric potential

r₁ is the first separation distance= 2.00×10−10 m

r₂ is the second  separation distance=  3.00×10−15 m

The electric potential generated by the proton at rest at the two points, using the formula:

Firstly the electric potential at loction 1

The electric potential at loction 2

The product of difference of electric potential and charge is defined as the workdone.

Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.

To learn more about electric potential work refer to the link.

brainly.com/question/12707371

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

where is the charge of the proton and is the potential difference between the final position and the initial position of the proton. To calculate this , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

where is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

Therefore, the work done is

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.