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masha68 [24]
4 years ago
14

Which object extends farther from the nucleus of the Milky Way galaxy?

Physics
1 answer:
sweet-ann [11.9K]4 years ago
4 0
The halo extends farther from the nucleus
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If a receiver is overly selective:
VMariaS [17]

Answer:

C) only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results.

Explanation:

Selectivity is the ability of a receiver to respond only to a specific signal on a wanted frequency and reject other signals nearby in frequency.

If a receiver is overly selective, only part of the bandwidth of the AM signal is amplified, causing some of the sideband information to be lost and distortion results. Whereas, if a receiver is underselective, the receiver can pick different signals on different frequencies at the same time.

7 0
3 years ago
Help ;-;
nasty-shy [4]

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

3 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. a little later it returns to its point of release. the ball
Lady_Fox [76]
<span>Final Velocity = Vf = 0 m/s --------------> (Vf = 0 because ball's speed at its max height is 0) Initial Velocity = Vi = ? Total time (upward & downward) = 8.0 seconds * Time upward = 4 seconds & ................( As time for ball upward & downward is equal ) * Time downward = 4 seconds.. Gravitational Acceleration = g = -9.8 m/s² Use Equation; Vf = Vi - gt 0 = Vi - 9.8 * 4 0 = Vi - 39.2 39.2 = Vi => Vi = Initial Velocity = 39.2 m/s</span>
7 0
3 years ago
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
3 years ago
Which type of biological molecule would contain fats?
Nana76 [90]

B

Explanation:

lipids contains fat

hope it helps

5 0
3 years ago
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