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3 years ago

Weight of a man is less at coal mine.Why?

Physics

2 answers:

gavmur [86]3 years ago

6 0

There is less gravity underground

dmitriy555 [2]3 years ago

5 0

If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.

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¿Qué significa que la investigación debe ser replicable? Analiza.

Inga [223]

Significa que en su investigación, debe proporcionar suficiente información para que otras personas que lean su investigación puedan hacer la investigación nuevamente.

4 0

3 years ago

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2\times 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=\dfrac{F}{A}$

$F=P\times A$

$F=2.2\times 10^5\times 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0

3 years ago

Which part of the electromagnetic spectrum is nearest to radio waves?

Sav [38]

Answer:

i believe it is the microwaves that is closest to the radio waves.

Explanation:

8 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa

Darya [45]

Answer:

x=0.53 $x10^{-3} m$

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1. $x10x^{-6} \frac{C}{m^{2}}$

ε=8.85 $x10^{-12} \frac{C^{2}}{n*m^{2}}$

$E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}$

Force of charge is

$F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N$

$F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s$

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

$x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m$

5 0

3 years ago

I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s

masha68 [24]

20/9.8 = 2.0 seconds. The ball stops after 2 seconds.

5 0

3 years ago