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brilliants [131]
4 years ago
10

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student hold

s a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?
Physics
1 answer:
gayaneshka [121]4 years ago
5 0
B. 60 cm 

  All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.
Hope this helps !
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If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
     TRUE

D) All of the above.
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4 0
3 years ago
Read 2 more answers
Consider the magnetic field (B) of a wire with a constant current (I). A Magnetic Field sensor is placed at a radius (r). Will t
iren [92.7K]

Answer:

No, the magnitude of the magnetic field won't change.

Explanation:

The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.

8 0
3 years ago
Two cars of the same mass have different velocities. Which car has more momentum?
never [62]

<u>Answer:</u>

<h3>The car with the higher velocity would have a more impactful momentum.</h3>
7 0
3 years ago
How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?
Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

Q = ms\Delta T

We know that

melting temperature of copper = 1085 degree C

Specific heat capacity of copper = 385 J/kg C

now we have

Q = 3(385)(1085 - 83)

Q = 1157310 J

Q = 1157.3 kJ

now in order to melt the copper we know the heat required is

Q = mL

here we know that

L = 205 kJ/kg

now from above formula

Q = 3(205) kJ

Q = 615 kJ

now total heat required will be

Q = 1157.3 kJ + 615 kJ

Q = 1772.3 kJ

As we know that

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6 0
3 years ago
If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o
Scilla [17]

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

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When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

8 0
4 years ago
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