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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
If a person is kicking there legs three times every four seconds what is the frequency.
zheka24 [161]

Answer:

f = 0.75 Hz

Explanation:

It is given that, a person is kicking there legs three times every four seconds. We need to find his frequency. The number of times an event is occurring is called frequency of an object. The frequency of a person is given by :

f=\dfrac{n}{T}\\\\f=\dfrac{3}{4}\\\\f=0.75\ Hz

So, the frequency of the person is 0.75 Hz.

3 0
3 years ago
What is the total resistance of a 3-ohm resistor and a 6-ohm resistor in parallel?
Fittoniya [83]
1/R=1/3+1/6
1/R=2/6+1/6=3/6
R=6/3=2ohms
Total R=2ohms
5 0
3 years ago
A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th
Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

5 0
3 years ago
In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl
irga5000 [103]

Answer:

1.5048\times 10^{-23}\ Am^2

Explanation:

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

r = Radius of circle = 5.7\times 10^{-11}\ m

v = Velocity of proton = 3.3\times 10^6\ m/s

Magnetic moment is given by

M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2

The magnetic moment associated with this motion is 1.5048\times 10^{-23}\ Am^2

5 0
3 years ago
The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci
nasty-shy [4]

Answer:

\theta = 25.3^\circ

Explanation:

The acceleration of the block can be found by the kinematics equations:

v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ

7 0
3 years ago
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