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Karo-lina-s [1.5K]
3 years ago
14

Describe sound and record​

Physics
2 answers:
Iteru [2.4K]3 years ago
8 0

Audible: a sound that is loud enough to hear

A record is a database entry that may contain one or more values

oksian1 [2.3K]3 years ago
3 0

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

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a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

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3 years ago
Does the sun sun traces shortest path across local sky on june solstice
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the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

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Answer:

Point c

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