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Sergio039 [100]
3 years ago
5

Unlike images produced by a convex spherical mirror, images produced by concave spherical mirrors

Physics
2 answers:
daser333 [38]3 years ago
7 0
I think it's B, I know it's not C
Olenka [21]3 years ago
6 0
<h3><u>Answer;</u></h3>

B. can be real or virtual depending on the object's placement.

<h3><u>Explanation;</u></h3>
  • Unlike images produced by a convex spherical mirror, images produced by concave spherical mirrors <em><u>can be real or virtual depending on the object's placement.</u></em>
  • <em><u>Concave mirror and convex mirror are two types of spherical mirrors, where a convex mirror has a reflecting surface that bulges outside while a concave mirror has a reflecting surface that bulges inwards.</u></em>
  • <em><u>Convex mirror forms images that are virtual and diminished, while a concave mirror forms an image that may either be real or virtual, diminished or enlarged depending on the position of the object from the mirror.</u></em>
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Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat
vichka [17]
The period of a pendulum is given by
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
\frac{T_m}{T_e} =  \frac{2 \pi  \sqrt{ \frac{L}{g_m} } }{2 \pi  \sqrt{ \frac{L}{g_e} } }=    \sqrt{ \frac{g_e}{g_m} }
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is g_e = 9.81 m/s^2 while on the Moon is g_m=1.63 m/s^2, the ratio between the period on the Moon and on Earth is
\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45

3 0
3 years ago
Friction _____.
quester [9]

Answer:

all above

Explanation:

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8 0
2 years ago
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels
Kryger [21]

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

T = \dfrac{v^2.m}{l}

v = \dfrac{l}{s}

v = \dfrac{7.2}{0.74}

v = 9.73 m/s

now, calculation of tension

T = \dfrac{9.73^2\times 0.46}{7.2}

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The tension in the cord is equal to 6.0 N.

7 0
3 years ago
What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

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3 years ago
A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati
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Gravitational potential energy  = 

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                =    (5.8 kg)  x  (9.8 m/s²)  x  (2.5 m)

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