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Volgvan
3 years ago
8

What is the average speed of particles of atoms at room temperature?

Physics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

300 meters per second. That's equal to about 670 miles per hour.

Explanation:

Not only are air particles incredibly small, they are always moving. And they move fast. At room temperature, they are going about 300 meters per second. That's equal to about 670 miles per hour.

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A measure of how easily current will pass through a material ?
Jlenok [28]

Answer:

It is Conductivity because it is the measure of the ease.

6 0
3 years ago
A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that
saveliy_v [14]

Answer:

The speed it reaches the bottom is

v=6.51m/s

Explanation:

Given: m=5.0kg, r=47cm\frac{1m}{100cm}=0.47m

Using the conservation of energy theorem

U_i=K_E+K_{ER}

m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2

v=r*w, I=\frac{1}{2}*m*r^2

m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2

m*g*h=\frac{3}{4}*m*r^2*w^2

g*h=\frac{3}{4}*r^2*w^2

Solve to w'

w^2=\frac{4*g*h}{3*r^2}

h=x*sin(30)=6.5m*sin(30)=3.25m

w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}

w=27.74rad/s

v=27.74rad/s*0.235m=6.51m/s

7 0
3 years ago
How did Orsted discover Electromagnetism?
Liula [17]

Answer:

Hans Christian Oersted began a new scientific epoch when he discovered that electricity and magnetism are linked. He showed by experiment that an electric current flowing through a wire could move a nearby magnet. The discovery of electromagnetism set the stage for the eventual development of our modern technology-based world.

Explanation:

7 0
3 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

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3 years ago
What is the importance of star ranting labelilling of an electrical appliance?​
Misha Larkins [42]

Answer:

The standard of efficiency varies by product category

these are called BEE star label and they show how much electricity the appliance consumes in year each appliance gets btw one and five stars with five stars meaning that it's extremely efficienct andis likely to keep your electricity bills in check

<em><u> </u></em><em><u>hop</u></em><em><u>e</u></em><em><u> it's</u></em><em><u> helpful</u></em>

5 0
3 years ago
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