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3 years ago

What is the magnitude of the force of gravity acting on a box that has a mass of 100 kilograms and is at sea level

Physics

2 answers:

vovikov84 [41]3 years ago

5 0

Answer:

Force of gravity, F = 980 N

Explanation:

Given that,

The mass of the box, m = 100 kilograms

We need to find the magnitude of the force of gravity acting on a box. It can be calculated using Newton's second law of motion which states the force is given by :

F = ma

a = acceleration

Here,

F = mg

g = acceleration due to gravity

$F=100\ kg\times 9.8\ m/s^2$

F = 980 N

Hence, the force acting on the object is 980 N.

jonny [76]3 years ago

3 0

The answer would be 981 newtons or 220.46 pounds.

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3 years ago

Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on

mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

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Therefore the change in the energy would be:

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3 years ago

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Mars2501 [29]

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3 years ago

How many photons are absorbed during the dental x-ray?

MariettaO [177]

<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
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8 0

3 years ago

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

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3 0

3 years ago