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aniked [119]
3 years ago
10

3. Atmospheric pressure is often given in units of millimeters of mercury. This refers to the height of a mercury column above t

he mercury's surface at the base of a barometer . The force exerted by the atmosphere on the surface of the mercury in the reservoir equals the weight of the mercury in the column . If the mercury column extends 760 mm above the mercury's surface in the reservoir , what is the atmosphere's pressure over the mercury ? Use 13.6*10^ 3 kg/m^ 3 for the density of mercury
Physics
1 answer:
gregori [183]3 years ago
6 0

Answer:

101397.16 pa

Explanation:

The pressure recorded will be equal to pgh

Where p = density of mercury = 13.6x10^3 kg/m^ 3

g = acceleration due to gravity 9.81 m/s^2

h = height of mercury in the column = 760 mm = 760x10^-3 m

Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa

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. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?
Hoochie [10]

The acceleration of the object is 6.7 m/s^2

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
3 years ago
?????Help please?????
Sindrei [870]

The correct selections are C, C, B, D, A, B, and A .

6 0
3 years ago
A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The str
alekssr [168]

Answer:

F = 5253.7 N

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

T = \frac{mv^2}{L}

now we have

68 = \frac{m(16.53^2)}{3.7}

now we have

68 = 73.8 m

m = 0.92 kg

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

F = \frac{mv^2}{r}

F = \frac{0.92(71.5^2)}{0.896}

F = 5253.7 N

8 0
3 years ago
Read 2 more answers
Why do areas north of the artic circle in tje northern hemisphere experience a polar day lasting for several months during summe
adell [148]
<span>The regioin is titled towqrd the Sun during polar day.  (C)

(The same exact thing happens in areas south of the Antarctic Circle 
in the southern hemisphere.  The only difference is that the whole thing
is spelled better in the South.)</span>
7 0
3 years ago
Read 2 more answers
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