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4 years ago

Can anyone help me with this

Physics

1 answer:

alukav5142 [94]4 years ago

3 0

There are four laws of thermodynamics.

Zeroeth Law of Thermodynamics: Two systems in thermal equilibrium with a third system are in thermal equilibrium to each other.

First Law of Thermodynamics: The change in a system's internal energy is equal to the difference between heat added to the system from its surroundings and work done by the system on its surroundings.

Second Law of Thermodynamics: It is impossible for a process to have as its sole result the transfer of heat from a cooler body to a hotter one.

Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero is exactly equal to zero.

You might be interested in

Ice of mass 5 g at 0 °C melts to water at 0 °C.

amid [387]

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

5 0

3 years ago

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

$m=180 kg$ the mass of the raft

$F=75 N$ the force applied by Sawyer

$V=2 m/s$ the raft's final speed

$V_{o}=0 m/s$ the raft's initial speed (assuming it starts from rest)

We have to find the time $t$

Well, according to Newton's second law of motion we have:

$F=m.a$ (1)

Where $a$ is the acceleration, which can be expressed as:

$a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}}$ (2)

Substituting (2) in (1):

$F=m\frac{V-V_{o}}{t-t_{o}}$ (3)

Where $t_{o}=0$

Isolating $t$ from (3):

$t=\frac{m(V-V_{o})}{F}$ (4)

$t=\frac{180 kg(2 m/s-0 m/s)}{75 N}$

Finally:

$t=4.8 s$

6 0

3 years ago

03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft

Flauer [41]

Answer:

a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,

e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

w = $\sqrt{\frac{k}{m} }$

a) find the period

angular velocity, frequency, and period are related

w = 2π f = 2π / T

f = 1 / T

T = 1 / f

T = 1 / 0.59

T = 1.69 s

b) the spring constant

w = 2π f

w = 2π 0.59

w = 3.70 rad / s

w² = k / m

k = w² m

k = 3.70² 0.060

k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

x = A cos (wt + Ф)

speed is defined by

v = $\frac{dx}{dt}$

v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

v = A w

v = 0.394 3.70

v = 1.46 feet/s

d) maximum acceleration

a = $\frac{dv}{dt}$

a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

a = A w²

a = 0.394 3.70²

a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

v = -A w sin (0 + Фi)

0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

x = A cos wt

x = 0.394 cos 3.70t

we substitute

0.1969 = 0.394 cos 370t

3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

3.70, t = 1.047

t = 1.047 / 3.70

t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

v = - Aw sin wt

v = - 0.394 3.70 sin 3.70 0.2826

v = - 1,319 ft / s

a = - A w² cos wt

a = - 0.394 3.70² cos 3.70 0.2826

a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

K = ½ m v²

let's slow down to the SI system

v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

K = ½ 0.060 0.402²

K = 4.8 10⁻³ J

U = ½ k x²

U = ½ 0.825 0.06²

U = 1.49 10⁻³ J

5 0

3 years ago

Bill and amy want to ride their bikes to school which is 14.4 kilometers away. It takes Amy 49 minutes to get to school and bill

Drupady [299]

3.4m/s

Explanation:

Given parameters:

Distance to school = 14.4km

Time taken by Amy = 49min

Time taken by bill = 20min after Amy = 20+49 = 69min

Unknown parameters:

How much faster is Amy's average speed = ?

Solution:

Average speed is the rate of change of total distance with total time taken.

Average speed = $\frac{total distance }{total time taken}$

convert units to meters and seconds

1000m = 1km

60s = 1min

Distance to school = 14.4 x 1000 = 14400m

Time taken by Amy = 49 x 60 = 2940s

Time taken by Bill = 69 x 60 = 4140s

Average speed of Amy = $\frac{14400}{2940}$ = 4.9m/s

Average speed of Bill = $\frac{4140}{2940}$ = 1.4m/s

Differences in speed = 4.9 - 1.5 = 3.4m/s

Amy was 3.4m/s faster than Bill

learn more:

Average speed brainly.com/question/8893949

#learnwithBrainly

5 0

4 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago