<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>
5 km northeast. Left and up would make northeast
Answer:
a) FE = 0.764FG
b) a = 2.30 m/s^2
Explanation:
a) To compare the gravitational and electric force over the particle you calculate the following ratio:
(1)
FE: electric force
FG: gravitational force
q: charge of the particle = 1.6*10^-19 C
g: gravitational acceleration = 9.8 m/s^2
E: electric field = 103N/C
m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg
You replace the values of all parameters in the equation (1):
Then, the gravitational force is 0.764 times the electric force on the particle
b)
The acceleration of the particle is obtained by using the second Newton law:
you replace the values of all variables:
hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.
Answer:
Volume = 1,015 acre-feet (Approx)
Explanation:
Given:
Rain = 1.7 in
Time = 30 min
Area = 29 km²
Find:
Volume in acre-feet
Computation:
1 km = 1,000 m
1 m = 3.28 feet
1 km² = 247.105 acre
d = 1.7 in = 1.7 / 12 = 0.14167 ft
Area = 29 × 247.105 = 7,166.045 acre
Volume = 7,166.045 acre × 0.14167 ft
Volume = 1,015 acre-feet (Approx)
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
Read more on waves here
brainly.com/question/25699025
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
