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Readme [11.4K]
2 years ago
12

An elevator is moving down with an acceleration of 3.36 m/s2.

Physics
1 answer:
sergeinik [125]2 years ago
5 0

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

  • From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
  • When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

\sf\longrightarrow Weight = mg - ma \\

\sf\longrightarrow Weight = m ( g - a ) \\

  • Mass of the man = 64.2 kg

\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\

\sf\longrightarrow Weight = 64.2 * 6.44 N\\

\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\

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Compared to the cells in a healthy body, Elisa’s cells are getting far fewer ____________________ molecules.
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Explanation:

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3 years ago
A spaceprobe has an 29.0 m length when measured at rest. What length
elixir [45]

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let L_0 be the length of the space-probe when measured at rest, and L be its length as observed by an observer moving at velocity v, then

(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }

Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

3 0
3 years ago
True or false gravitational pull decreases with an increase of distance between two objects​
ohaa [14]

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

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3 years ago
A 5kg bag falls a verticle height of 10m before hitting the ground.
g100num [7]

Answer:

u = 7m {s}^{ - 1}

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

me(1) = me(2)

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

we divide both sides of the equation with mass to make the math easier.

9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

7 0
3 years ago
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