CALLING ALL BRAINLIERS THE FIRST PERSON TO ANSWER GETS A BRAINLIEST ANNNNDDD 30 POINTS BUT THE ANSWER HAS GOT TO BE RIGHT.

When considering the gas laws, the kelvin temperature scale must be used. the reason for this is that the kelvin scale is direct

suter [353]

.;;jmbbbvvb nhjmhnjhmnhjmgmjhyg gt vfdxg fbrdhtggdfse

7 0

3 years ago

Two cars are traveling on a desert road. After 5.0 seconds they are side by side at the next telephone pole.The distance between

stira [4]

The question looks incomplete, but according to the information given above seem like they have identical journeys.
a. the displacement of car A - 65.5 m
b. the displacement of car B - 65.5 m
c. average velociry of A $65.5 / 5 = 13.1 m/s$
d. the average velocity of car B has the same.

8 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5

Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

8 0

2 years ago

A motorcycle moving 18.8 m/s has 57800 J of KE. What is its mass?

Brums [2.3K]

Answer:

m = 327.07 kg

Explanation:

Given that,

Kinetic energy of a motorcycle, E = 57800 J

Velocity of the motorcycle, v = 18.8 m/s

We need to find the mass of the motorcycle. The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

m is mass

$m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg$

So, the mass of the motorcycle is 327.07 kg.

8 0

2 years ago