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ololo11 [35]
3 years ago
9

Uses of concave lens​

Physics
1 answer:
natita [175]3 years ago
3 0

Answer:

Concave Lens Uses. Telescope and Binoculars Spectacles Lasers Cameras FlashlightsPeepholes. ...

Used in telescope and binoculars. ...

Concave lens used in glasses. ...

Uses of concave lens in lasers. ...

Use of concave lens in cameras. ...

Used in flashlights. ...

Concave lens used in peepholes.

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to
nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket V=10\ gallon

Time taken to fill the bucket t=50\ s

so volume flow rate is \dot{V}=\frac{10}{50}=0.2\ gal/s

1 gal is equivalent to 0.133\ ft^3

\dot{V}=0.0267\ ft^3/s

mass flow rate \dot{m}=\rho \times \dot{V}

\dot{m}=62.4\times 0.0267

\dot{m}=1.668\ lbs

(b)Average velocity through nozzle exit

\dot{V}=Av_{avg}

v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}

v_{avg}=49.51\ ft/s

8 0
4 years ago
A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
3 years ago
A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?
prohojiy [21]

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0
4 years ago
An electric charge, A, is placed carefully between two other charges, B and C, and experiences no net electric force. Do B
Brrunno [24]

Answer:

I do not have  enough information to tell

Explanation:

This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.

5 0
3 years ago
How do you do this problem?
mojhsa [17]

Answer:

C

Explanation:

When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.

Now C has a charge of +12

When A and C come together they each have an equal share of that 12, so each of them has 6

So the answer is

A       B       C

6        0       6

which is C

4 0
3 years ago
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