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amm1812
4 years ago
6

You don't have any matches, so you want to focus sunlight to ignite a pile of twigs for a campfire. Which mirror would be best t

o use? A.convex mirror B.concave mirror C.flat mirror D.plane mirror
Physics
2 answers:
Anna35 [415]4 years ago
4 0
Out of that list, the concave mirror is the only item that can concentrate sunlight and heat into a small area. But if you could get ahold of a convex lens, that would be even better.
Lera25 [3.4K]4 years ago
3 0

Answer: Concave mirror would be best to use.

Explanation :

To focus sunlight to ignite a pile of twigs for a campfire, a concave mirror should be used.

When the rays of light coming from infinity parallel to principal axis in front of a concave mirror, all the rays get concentrated at the focus of the mirror. That's why the concave mirror is also known as converging.

In this case, a concave mirror should be used so as to converge all the rays at a point.

So, the correct option is (b).        

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A G note has a frequency of 783.99 Hz. Calculate its wavelength in meters.
Alla [95]
The answer is 0.434 m
8 0
3 years ago
Read 2 more answers
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
tino4ka555 [31]

Answer:

(a) Negative Q

(b) Positive Q

Explanation:

Charge is the inherent property of matter due to the transference of electrons.

There are three methods of charging a body.

(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.  

(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.

(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.  

(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.

(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.

7 0
3 years ago
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9
rusak2 [61]

Answer:

3.9 × 10^7 J

Explanation:

Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Solution

Since the tank is half full, the height = 2.5m

Pressure = density × gravity × height

Pressure = 900 × 9.8 × 2.5

Pressure = 22050 Pascal

The cross sectional area of the pump will be area of a circle.

A = πr^2

A = π × 15^2

A = 706.858 m^2

Using the formula

Density = mass/volume

Mass = density × volume

Mass = 900 × 706.86 × 2.5

Mass = 1590.435

Energy = mgh

Energy = 1590.435 × 9.8 × 2.5

Energy = 38965657.8 J

Since the work done = energy

Therefore, the work done = 3.9 × 10^7 J

7 0
3 years ago
A diamond sparkles more than a glass cut to similar shapes,why?​
8090 [49]

Answer:

because when you shape glass it's not a shiny as diamond

4 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.270 rev/s . The magnitude
Sauron [17]

Answer:

A)0.456 rev/s

B) 0.075 rev

C) 1.06 m/s

D) 3.7 m/s²

Explanation:

Initial angular velocity = \omega _{o} = 0.270 rev/s =  1.6965 rad/s

Angular acceleration = α = 0.899 rev/s/s = 5.64858 rad/s/s

Radius = r = 0.74/2 = 0.37 m

Time = t = 0.207 s

A) Final angular velocity is given by the equation

=\omega_{f}=\omega _{o}+ \alpha t

=1.6965 + (5.64858)(0.207) = 2.87 rad/s = 0.456 rev/s

B) No. of revolutions is calculated using the equation

\theta =\omega_{o}t + 1/2 \alpha t^2 = (1.6965 \times 0.207 )+ 1/2 (5.64858)(0.207)^2= 0.472 rad  = 0.075 revolutions

C) v = r ω =(0.37)(2.87) = 1.06 m/s

D) Centripetal acceleration = a_{c}=r\omega_{f} ^{2} = (0.37) (2.87)² =3.05 m/s²                                

Tangential acceleration =a_{t}=r \alpha

= ( 0.37)(5.64858) = 2.08997 m/s²

Resultant acceleration = a = \sqrt{a_{t}^{2} + a_{c}^{2

= 3.697 m/s²

8 0
4 years ago
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