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ANEK [815]
2 years ago
5

What is the power of a diverging lens of focal length 0.4 m

Physics
1 answer:
romanna [79]2 years ago
4 0

Answer:

-2.5 is the answer to your question

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Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum
loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45

The minimum sum of the resultant vector,

R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10

So, the magnitude of this sum either 45 or -10.

6 0
2 years ago
Which of the following is a scalar quantity?
neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0
2 years ago
Read 2 more answers
What is the angle of deviation in a plane mirror at normal incidence?​
Tems11 [23]

Answer:

The deviation of a mirror is equal to twice the angle of incidence.The total angle between the straight-line path and the reflected ray is twice the angle of incidence. This is called the deviation of the light and measures the angle at which the light has strayed from its initial straight-line path.

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

7 0
3 years ago
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
On a cold winters day if you left a cup of water sitting outside it could freeze heat is transferred out of the water describe t
valkas [14]
The water molecules would slow down, and as they slow down, the heat created from their movement would cease.
4 0
3 years ago
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