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2 years ago

Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?

Physics

1 answer:

morpeh [17]2 years ago

3 0

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays, parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

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Kathmandu lies at high altitude than biratnagar from sea level.Where does an object has more weight between two places?Give reas

Alla [95]

Answer:

Kathmandu

Explanation:

As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.

8 0

2 years ago

A Roller Derby Exhibition recently came to town. They packed the gym for twoconsecutive weekend nights at South's field house. O

Alla [95]

Answer:

14.4 m/s

Explanation:

mass of Anna (Ma) = 68 kg

speed of Anna (Va) = 17 m/s

mass of SandraDay (Ms) = 76 kg

speed of SandraDay (Vs) = 12 m/s

We can find their speed (V) immediately after collision from the conservation of momentum where

(Ma x Va) + (Ms + Vs) = (Ma + Ms) x V

where V = speed immediately after collision

(68 x 17) + (76 + 12) = (68 + 76) x V

2068 = 144 V

V = 2068 / 144 = 14.4 m/s

8 0

2 years ago

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0

2 years ago

A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract

Ivan

Answer:

Explanation:

Let v be the linear velocity , ω be the angular velocity and I be the moment of inertia of the the puck.

Kinetic energy ( linear ) = 1/2 mv²

Rotational kinetic energy = 1/2 I ω²

I = 1/2 m r² ( m and r be the mass and radius of the puck )

Rotational kinetic energy = 1/2 x1/2 m r² ω²

= 1/4 m v² ( v = r ω )

Total energy

= Kinetic energy ( linear ) + Rotational kinetic energy

= 1/2 mv² + 1/4 m v²

= 3/4 mv²

rotational K E / Total K E = 1/4 m v² / 3/4 mv²

= 1 /3

So 1 /3 rd of total energy is rotational K E.

3 0

2 years ago

The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba

snow_lady [41]

Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10

Also,
R₂*i = 9.5 (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0

2 years ago

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