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mariarad [96]
3 years ago
14

Can concave lens be used to make hand lens .why?​

Physics
1 answer:
mario62 [17]3 years ago
8 0

Answer:

No concave lens cannot be used to make hand lens because in hand Lens we use convex lens so as to converge the rays of the light After refraction and helps to produce a magnified image of an object.

Hope it will help you :)❤

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A soccer ball is rolling across a soccer field. The system consists of the
neonofarm [45]

Answer:

NEGATIVE

Explanation:

5 0
3 years ago
Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

Proven

5 0
2 years ago
For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m
Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

\Sigma F = F-R = 0 (Eq. 1)

Where:

F - Forward trust, measured in newtons.

R - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: (F = 6.85\times 10^{5}\,N)

R = F

R = 6.85\times 10^{5}\,N

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

4 0
3 years ago
Does wind have atoms and molecules in it?
Likurg_2 [28]
<span>
Of course.  Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc.  That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>


4 0
3 years ago
Read 2 more answers
The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
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