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artcher [175]
2 years ago
5

If the column of mercury in a barometer stands at 72.6 cm, what is the atmospheric pressure?

Physics
1 answer:
lakkis [162]2 years ago
6 0

Answer:pressure = density * acceleration due to gravity * height

H=72.6cm= 0.726m

P=0.726*13.6*10^3*9.8

P=96761.28Pa

Explanation:

You might be interested in
Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what
yaroslaw [1]

Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence \lambda= 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

t=\frac{\lambda }{4n} , here t is thickness, \lambda is wavelength and n is refractive index

Putting all values t=\frac{600}{4\times 1.5}=100nm

So minimum thickness will be 100 nm

8 0
3 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
bija089 [108]

This question is incomplete, the complete question is;

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.

At what rate is the magnetic field changing?

Answer:

the magnetic field changing at the rate of 9.33 m T/s

Explanation:

Given the data in the question;

Electric field E = 7 mV/m

radius r = 1.5 m

Now, from Faraday law of induction;

∫E.dl = d∅/dt

E∫dl = A( dB/dt )

E( 2πr ) = πr² ( dB/dt )

( 0.007 ) = (r/2) ( dB/dt )

( 0.007 ) = 0.75 ( dB/dt )

dB/dt = 0.007 / 0.75

dB/dt = 0.00933 T/s

dB/dt = ( 0.00933 × 1000) m T/s

dB/dt = 9.33 m T/s

Therefore, the magnetic field changing at the rate of 9.33 m T/s

5 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
The process of deposition causes The process of deposition causes
docker41 [41]
A land form or land mass to be created over a long period of time
7 0
3 years ago
Read 2 more answers
1. Bone has a Young’s modulus of about
Blababa [14]

#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

5 0
3 years ago
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