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sertanlavr [38]
2 years ago
13

Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by

the unaided eye. c) A virtual image cannot be photographed. d) A real image must be erect. e) Mirrors always produce real images because they reflect light.
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

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An astronaut on a space walk floats a little too far away
Lady bird [3.3K]

Answer:

He can throw it away from himself.

Explanation:

Newtons Third Law says that everything has an equal, yet opposite reaction on other objects.

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1 year ago
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An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .
miss Akunina [59]

Answer:

The airplane should release the parcel 6.7*10^2 m before reaching the island

Explanation:

The height of the plane is y_0=100m, and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

y=y_0 - \frac{gt^2}{2}    [1]

And the distance X is

x = V.t     [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

X = V . \sqrt{\frac{2y_0}{g}}

Using the given values:

x = 150 m/s  \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}

x = 6.7*10^2 m

4 0
2 years ago
An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of
satela [25.4K]
<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>
8 0
2 years ago
Select True or False for the following statements about Heisenberg's Uncertainty Principle. True False It is not possible to mea
sveticcg [70]

Answer:

Statement 1) False

Statement 2) False

Statement 3) True

Explanation:

The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

P\cdot\delta x\geq \frac{h}{4\pi }

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.

Statement 1 is false since measurement of x and y positions has no relation to uncertainty.

Statement 2 is false as both the momentum components can be measured with 100% precision.

Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.

6 0
1 year ago
The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the
otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

8 0
2 years ago
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