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Free_Kalibri [48]
3 years ago
15

A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it

Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

(a) convex mirror

(b) virtual and magnified

(c) 23.3 cm

Explanation:

The having mirror is convex mirror.

distance of object, u = - 20 cm

magnification, m = 1.4

(a) As the image is magnified and virtual , so the mirror is convex in nature.

(b) The image is virtual and magnified.

(c) Let the distance of image is v.

Use the formula of magnification.

m =-\frac{v}{u}\\1.4=-\frac{v}{-20}\\v =28 cm

Use the mirror equation, let the focal length is f.

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{f}=\frac{1}{28}+\frac{1}{20}\\\frac{1}{f}=\frac{28+20}{560}\\f=11.67cm

Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm

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Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =
Crank

Answer:

Fnet = 12 N

Explanation:

Force on a point charge due to another point charge = kq1q2 / d^2

Force on +32uC = due to + 20uC + due to -60uC

where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2

Net Force =

= \frac{1}{4\pi \epsilon _0} [\frac{32 \times 10^-^6 \times60\times10^-^6}{(60/100)^2}-\frac{32 \times 10^-^6 \times20\times10^-^6}{ (40/100)^2}  ]

F_{net}=9 \times10^9\times 10^-^1^2[\frac{32\times60\times10^4}{60\times60} -\frac{32\times20\times10^4}{40\times40} ]

=90[32(\frac{80-60}{60\times 80} )]\\\\=90\times32\times0.004167\\\\=12N

Fnet = 12 N

8 0
4 years ago
On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe
polet [3.4K]
The period of a simple pendulum is given by
T= 2 \pi  \sqrt{ \frac{L}{g} }
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
T_e= 2 \pi \sqrt{ \frac{L}{g_e} }
where g_e is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
T_m= 2 \pi \sqrt{ \frac{L}{g_m} }
where g_m is the acceleration of gravity on the Moon. 

If we do the ratio of the two periods, we get
\frac{T_m}{T_e} =  \sqrt{ \frac{g_e}{g_m} }
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write g_e = 6 g_m and we can rewrite the previous ratio as
\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }=  \sqrt{6}

so the period of the pendulum on the Moon is
T_m =  \sqrt{6}  T_e =  \sqrt{6} (1.6 s)=3.9 s
8 0
4 years ago
Pls i need this now
dmitriy555 [2]
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6 0
3 years ago
An offshore rig begins to spill oil in a circular patch centered on the rig. If the radius ofthe oil spill increases at a rate o
Tresset [83]

Answer:

Explanation:

The area of a circle is given b

A=πr²

Then,

dA/dr= 2πr

Also give that,

dr/dt=30m/hr, r=100m

We want to find dA/dt

dA/dt=dA/dr×dr/dt

dA/dt= 2πr × 30

dA/dt= 2 ×π ×100×30

dA/dt=18849.56m²/s.

The rate of the area is 18849.56m²/s

7 0
3 years ago
Read 2 more answers
20 PTS!
klemol [59]
  • its acceleration is zeo

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

8 0
3 years ago
Read 2 more answers
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