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natka813 [3]
3 years ago
14

An astronaut uses a Body Mass Measurement Device to measure her mass. Part A If the force constant of the spring is 3200 N/m, he

r mass is 65 kg, and the amplitude of her oscillation is 1.7 cm, what is her maximum speed during the measurement
Physics
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

v = 0.119 m/s

Explanation:

Given that,

Force constant, k = 3200 N/m

Mass, m = 65 kg

Amplitude of her oscillation is 1.7 cm i.e. 0.017 m

We need to find her maximum speed during the measurement. Let it is v. It is given by the formula as follows :

v=\sqrt{\dfrac{k}{m}}A

Where,

A is the amplitude

v=\sqrt{\dfrac{3200}{65}}\times 0.017 \\\\v=0.119\ m/s

So, the maximum speed is 0.119 m/s.

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HI PLEASE HELP!
butalik [34]

Answer:

Chemical reactions occur slower at lower temperatures and faster at higher temperatures. When you put a glow stick in cold water, the chemical reaction slows down but will last for a longer period of time. When you put a glow stick in hot water, the reaction speeds up but will be over quicker.

Explanation:

(self-explanatory)

I hope this helps, have a nice day.

7 0
3 years ago
When the north pole of a magnet is moved towards the center of a loop of wire containing a galvanometer, the needle of the galva
andriy [413]

Answer:

Moving the magnet away from the center of the loop with its south pole facing the center of the loop.

Explanation:

Electromagnetic induction is due to a rapidly changing magnetic field, or loop area. The poles of the magnet induce current in the loop but in the opposite direction, depending on the direction of their relative motion. An approaching north pole will induce an anticlockwise current in the loop, while an approaching south pole will do the reverse. To get the galvanometer to flicker in the same direction as of that when the north pole was approaching, we move the magnet away from the center of the loop with its south pole facing the center of the loop.

6 0
3 years ago
f a stadium pays $11,000 for labor and $7,000 for parking, what would the stadium's parking revenue have to be if the stadium is
QveST [7]

Answer:

Explanation:

a

7 0
3 years ago
Read 2 more answers
A small 17 kilogram canoe is floating downriver at a speed of 3 m/s. What is the canoe's kinetic energy?
aksik [14]
KE=1/2mv^2

So from the given information me can plug in the values and find the KE or Kinetic Energy for the canoe;

KE=1/2(17kg)(3m/s)^2
KE=8.5kg*9m^2s^2= 76.5J of Kinetic Energy

Hope this helps, any questions please just ask. Thank you.
7 0
4 years ago
Read 2 more answers
Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
JulijaS [17]

Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

4 0
3 years ago
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