Ask question

Ask question

All categories

3 years ago

8

In which labeled portion of the curve would you use the heat of vaporization to calculate the heat absorbed? (image attached ins

ide!)
1, 2, 3, 4, or 5

**not sure :( Thank you!!

1 answer:

sergij07 [2.7K]3 years ago

3 0

<span>In the labeled portion of the curve ,you use the heat of vaporization to calculate the heat absorbed in the 4th portion. It is indicated in the picture that it is the region where vaporization occurs, that is why you need to consider this portion to calculate.</span>

You might be interested in

On Earth, a kangaroo jumping will eventually return to ground due to the unbalanced force of gravity. What law does this illustr

Kryger [21]

Newton’s law is the answer

6 0

3 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

3 years ago

The current in a circuit is tripled by connecting a 580 resistor in parallel with the resistance of the circuit. Determine the r

const2013 [10]

Answer:

1160 ohm

Explanation:

We are given that

R'=580 ohm

Current=3 I

We have to find the resistance of the circuit.

Let R be the resistance of circuit.

In parallel

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Using the formula

$\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}$

$R_{eq}=\frac{580R}{580+R}$

In parallel combination,Potential difference across each resistance remains same.

$V=IR$

Using the formula

$IR=3IR_{eq}$

$IR=3I\times \frac{580R}{580+R}$

$580+R=3\times 580=1740$

$R=1740-580=1160\Omega$

6 0

3 years ago

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of th

Tamiku [17]

Answer:

2.124 kg of water

Explanation:

height of the falls is about 48 meters.

Mass of water needed is 1kg = 1000g

Power needed is 106 watts.

The amount of energy in 106 watts in one sec is 106 joules.

To calculate the energy of the 1kg falling water = Mgh

Energy = 1000*9.81*48

Energy = 470880 joules.

1 megawatt is = 1000000watts

The kilogram of water needed is 1000000/470880 = 2.124 kg of water

3 0

3 years ago

A man is traveling from the back of a boat to the front of the boat at 2.0 m/s while the boat itself is traveling at 12.0 m/s to

Hatshy [7]

Based on the relative velocity of the man with respect to the boat and the dock:

Distance covered in 4.0 seconds relative to the boat = 8 m
Distance covered in 4.0 seconds relative to the dock = 56 m

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of a body relative to another body which serves as a reference point.

Relative velocity is a vector.

Considering the velocity of the man and the boat:

The relative velocity of the man with respect to the boat = 2.0 m/s

Distance covered in 4.0 seconds relative to the boat = 2.0 m/s * 4.0 s

Distance moved = 8 m

Relative velocity of the man with respect to the dock = 12 + 2 = 14 m/s

Distance covered in 4.0 seconds relative to the dock = 14.0 m/s * 4.0 s

Distance moved = 56 m

In conclusion, the relative velocity is velocity with respect to a reference point.

Learn more about relative velocity at: brainly.com/question/24337516

#SPJ1

5 0

1 year ago