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allsm [11]
3 years ago
12

You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

ation of the elevator
Physics
1 answer:
lbvjy [14]3 years ago
4 0
In a stationary situation, the weight of person is
W=mg=(62 kg)(9.81 m/s^2) = 608.2 N
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
\sum F = ma
W-W'=ma
where a is the acceleration of the elevator. If we solve for a, we find
a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
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If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?
Licemer1 [7]
<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

         Number of dribbles in 1 second = 1.77

         \texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s

Now we need to find how long does it take for you to complete 12 dribbles.

         Time taken for 12 dribbles = 12 x Time taken for 1 dribble

         Time taken for 12 dribbles = 12 x 0.565

         Time taken for 12 dribbles = 6.78 seconds      

It takes 6.78 seconds to complete 12 dribbles.  

8 0
2 years ago
5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-
Airida [17]
D. 980, this is the best answer because 35 x 7 is 980 :)

3 0
2 years ago
What is Force &amp; how many types of Force express with example <br><br><br><br>​
ehidna [41]

Answer:

The force is pull or push acting on the body which tends to change its state of rest or of motion is called force.

There are two types of force:

1.Contact force

2. Non-Contact

7 0
3 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l
lianna [129]
For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40 

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.
3 0
2 years ago
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