Answer:
Walking’ - If a handball player takes more than three steps without dribbling (bouncing the ball) or holds the ball for more than 3 seconds without bouncing it, shooting or passing, then that is deemed ‘walking' and possession is lost.
'Double dribble’ - Handball players cannot receive the ball and bounce it, then hold the ball, and bounce it again. This is termed ‘double dribble’ and is against the rules.
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Answer:
<u>For M84:</u>
M = 590.7 * 10³⁶ kg
<u>For M87:</u>
M = 2307.46 * 10³⁶ kg
Explanation:
1 parsec, pc = 3.08 * 10¹⁶ m
The equation of the orbit speed can be used to calculate the doppler velocity:
making m the subject of the formula in the equation above to calculate the mass of the black hole:
.............(1)
<u>For M84:</u>
r = 8 pc = 8 * 3.08 * 10¹⁶
r = 24.64 * 10¹⁶ m
v = 400 km/s = 4 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)
M = 590.7 * 10³⁶ kg
<u>For M87:</u>
r = 20 pc = 20 * 3.08 * 10¹⁶
r = 61.6* 10¹⁶ m
v = 500 km/s = 5 * 10⁵ m/s
G = 6.674 * 10⁻¹¹ m³/kgs²
Substituting these values into equation (1)
M = 2307.46 * 10³⁶ kg
The mass of the black hole in the galaxies is measured using the doppler shift.
The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.
Answer:
Q = 5267J
Explanation:
Specific heat capacity of copper (S) = 0.377 J/g·°C.
Q = MSΔT
ΔT = T2 - T1
ΔT=49.8 - 22.3 = 27.5C
Q = change in energy = ?
M = mass of substance =508g
Q = (508g) * (0.377 J/g·°C) * (27.5C)
Q= 5266.69J
Approximately, Q = 5267J
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
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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)?
