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nexus9112 [7]
3 years ago
5

If the magnification produced by a lens has a negative value, the image will be

Physics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

I think it's B, not quite sure tho.

damaskus [11]3 years ago
5 0
I believe the answer is A. Virtual and inverted.
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Which property of the sound wave changes when the volume is increased on a radio?.
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amplitude

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When we increase the volume of a sound, the amplitude of the wave is increased. The frequency of a wave is related to its pitch. If the pitch is high, then the frequency of the wave is high. This means that the wave will looked squashed on an oscilloscope trace.

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In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa
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tan^-1 (6.2/22)= 15.7º
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Astronomers define the __________ as all of space and everything in it. It is enormous, almost beyond imagination. Question 2 op
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3 years ago
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Morgarella [4.7K]

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Generally, magnets are attracted to objects that are made of the metals iron, nickel, or cobalt. These materials are called ferromagnetic materials. ... When all or most of the domains are aligned in the same direction, the whole object becomes magnetized in that direction and becomes a magnet.

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5 0
2 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
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