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2 years ago

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Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau

se . . .” format and be sure to answer the lesson question: "How is potential energy converted to thermal energy in a system?”

1 answer:

IgorLugansk [536]2 years ago

4 0

If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.

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The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

gregori [183]

The answers are:

a) $Work=125,923.61J$

b) $Power=1.46watt$

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

$1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}$

Also, we need to find the mass given the density of the blood.

$1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg$

Now, calculating how much work does the heart do in a day, we have:

$Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J$

Then, calculating what is the power output and its horsepower, we have:

$Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt$

Have a nice day!

7 0

2 years ago

You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O

alexira [117]

Answer:

v_wind = 101.46 km / h , θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

v = d / t

v = 750 / 3.14

v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and $v_{y}$

Let's use trigonometry to find the components of the plane's speed

cos θ = v_N / v

sin θ = v_W / v

v_N = v cos θ

v_W = v sin θ

let's calculate

V _N = 238.85 cos 22 = 221.46 km / h

v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

vₓ - v_W = 0

vₓ = v-w

vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

v_total = d / t

v_total = 750 / 4.32

v_total = 173.61 km / h

This is the final speed of the plane, which can be written

v_total = v_n - vy

vy = v_n - v_total

vy = 221.46 - 173.61

vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

v_wind² = vₓ² + vy²

v_wind = √ (89.47² + 47.85²)

v_wind = 101.46 km / h

the address will then be found using trigonometry

θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹1 (47.85 / 89.47)

θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

90- θtea = 90- 28.2

θ = 61.8

East of South

5 0

1 year ago

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

5 0

1 year ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

1 year ago

Using the right-hand rule from your lessons, determine the directions of the electrical current and magnetic field of the electr

aliya0001 [1]

Answer:

Hello there use something that looks like this

Explanation:

This is an accurate representation of something you are working on!

As you can see the wire and the core are represented on the left and is showing how it can be represented on your right hand and how they are similar!

5 0

2 years ago