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aivan3 [116]
2 years ago
9

Suppose the door of a room makes an airtight but frictionless fit in its frame. Do you think you could open the door if the air

pressure on one side were standard atmospheric pressure and the air pressure on the other side differed from standard by 1%? Explain.
Physics
1 answer:
Amiraneli [1.4K]2 years ago
3 0

Answer:

Yes, it could be opened

Explanation:

Pressure is force per unit area

 P =\frac{F}{A}

Since there is a difference in air pressure on both sides, this difference when multiplied by the area of the floor will create a force which could open the door, if this force is complimented with a greater applied force.

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16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.
miskamm [114]

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is  

\displaystyle E_1=\frac{mv^2}{2}

\displaystyle E_1=\frac{7,500\ 20^2}{2}

E_1=1,500,000\ Joule

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

U=mgh=(7,500)(9.8)(10)=735,000\ Joule

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule

7 0
2 years ago
In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?
faust18 [17]
Below are the choices that can be found in the other sources:

A. diffraction 
<span>B. refraction </span>
<span>C. reflection </span>
<span>D. transmission
</span>
The answer is diffraction. It means that <span>the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.</span>
8 0
2 years ago
The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
2 years ago
Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No
Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field  

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns  = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0
2 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
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