Ask question

Ask question

All categories

3 years ago

8

"A simple heat engine works by extracting heat from the hot reservoir, which is then converted into mechanical work, then unconv

erted heat is exhausted to the atmosphere"
True or false

1 answer:

Elanso [62]3 years ago

6 0

Umm no se nada perón

You might be interested in

Two trains approach each other on parallel tracks. Each has a speed of 95 kph relative to the ground. If they are initially 8.5

aliya0001 [1]

Answer: 2.83 minutes

Explanation:

It is understood that trains are approaching. That is, they have speeds of equal magnitude but opposite. When train A travels x meters northbound, then train B travels the same distance southbound.

Therefore trains approach at a speed of:

$V = 95 +95 = 190\ km / h$

Then:

$V = \frac{x}{t}\\\\t = \frac{x}{V}$

Where x is the distance between the trains

$x = 8.5\ km$

So the time in which both trains meet is:

$t =\frac{8.5\ km}{180\ \frac{km}{h}}\\\\t=0.04722\ hours$

This is:

$0.04722*\frac{60\ minutes}{1\ hour} = 2.83\ minutes$

<em />

<em>How long will it be before they reach one another ?</em>

<h3>2.83 minutes</h3>

3 0

3 years ago

Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a

kodGreya [7K]

Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

-Tₓ = m aₓ

Y Axisy

$T_{y}$ –W = m a_{y}

Let's use trigonometry to find the components of force

sin 21.5 = Tₓ / T

cos 21.5 = T_{y} / T

Tₓ = T sin 21.5

T_{y} = T cos 21.5

Tₓ = 768 sin 21.5 = 281.47 N

T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y Axis F_net = 714.56 - 825 = -110.44 N

F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

F = √ (Fₓ² + T_{y}²)

F = √ (281.47² + 110.44²)

F = 281.70 N

d) Let's use trigonometry

tan θ = F_{y} / Fₓ

θ = tan⁻¹ F_{y} / Fₓ

θ = tan⁻¹ (-110.44 / 281.47)

θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

θ = 360 - 21.42

θ = 338.58º

Acceleration

X axis

Tₓ = m aₓ

aₓ = Tₓ / m

The mass of Tarzan is

m = W / g

m = 825 / 9.8 = 84.18 kg

aₓ = 281.47 / 84.18

aₓ = -3.34 m / s2

Y Axis

T_{y}-W = m a_{y}

a_{y} = (T_{y} -W) / m

a_{y} = (714.56-825) / 84.18

a_{y} = - 1,312 m / s²

Acceleration Module

a = √ aₓ² + a_{y}²

a = √ (3.34² +1.312²)

a = 3,588 m / s²

The angle

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-1312 / -3.34)

θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

θ = 180 + 21.45

θ = 201.45º

3 0

3 years ago

How would i solve these problems, nobody in my class understands and there is a substitute

DiKsa [7]

1) X-component: 568.5 N, Y-component: 511.9 N

2) The horizontal force is 86.6 N and the resulting acceleration is $0.87 m/s^2$

Explanation:

1)

In this first part of the problem, we have to resolve the force into its two components, along the x and the y direction.

The two components are given by:

$F_x = F cos \theta\\F_y = F sin \theta$

where

F = 765 N is the magnitude of the force

$\theta=42.0^{\circ}$ is the angle of the force with the horizontal

Substituting, we find:

Horizontal component: $F_x = (765)(cos 42.0^{\circ})=568.5 N$
Vertical component: $F_y = (765)(sin 42.0^{\circ})=511.9 N$

2)

First of all, we have to find the horizontal component of the pulling force, which is given by

$F_x = F cos \theta$

where

F = 100 N is the magnitude of the pulling force

$\theta=30.0^{\circ}$ is the direction of the force with the horizontal

Substituting,

$F_x = (100)(cos 30^{\circ})=86.6 N$

Now we can find the acceleration of the wagon by using Newton's second law:

$F_x = ma_x$

where

$F_x = 86.6 N$ is the net force in the horizontal direction

m = 100 kg is the mass of the wagon

$a_x$ is the acceleration in the horizontal direction

Solving for $a_x$ , we find

$a_x = \frac{F_x}{m}=\frac{86.6}{100}=0.87 m/s^2$

Learn more about vector components:

brainly.com/question/2678571

And about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t

Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N

Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.

8 0

3 years ago

The distance between the centers of the wheels of a motorcycle is 156 cm. The center of mass of the motorcycle, including the ri

Drupady [299]

Answer:

a = 9.86 m/s²

Explanation:

given,

distance between the centers of wheel = 156 cm

center of mass of motorcycle including rider = 77.5 cm

horizontal acceleration of motor cycle = ?

now,

The moment created by the wheels must equal the moment created by gravity.

take moment about wheel as it touches the ground, here we will take horizontal distance between them.

then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.

now equating both the moment

m g d = F h

d is the horizontal distance

h is the vertical distance

m g d = m a h

term of mass get eliminated

g d = a h

so,

$a = \dfrac{g\ d}{h}$

$a = \dfrac{9.8\times 0.78}{0.775}$

a = 9.86 m/s²

4 0

3 years ago