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aniked [119]
3 years ago
5

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

value 3.8 A ten times in a time interval of 0.17 s. What is the period T with which the current at the wall socket changes?
Physics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A  ten times.  

So there must have been 10 cycles (10 periods) in 0.17s.   let 'T'  be the period:

T=\frac{t}{n}

t is the total time interval

n is the number of oscillations

T=\frac{0.17}{10}

10T = 0.17

T = 0.17/10 = 0.017s

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A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
3 years ago
Suppose you have thrown a rock nearly straight up at a coconut in a palm tree,and the rock misses on the way up but hits the coc
Margarita [4]
Most likely it would dislodge the coconut on the way down due to gravity because on the way up the gravity would slow down the rock but on they down the gravity pulls the rock
5 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
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