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zmey [24]
3 years ago
5

A tank of water is 4m deep. How deep does it appear when seen normally?​

Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

The tank appears to be 3 m deep only. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
andrew-mc [135]

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = \frac{100}{220} A

Case 2

P = V * I

P = 110 * \frac{100}{220}

P = 50 W

I think the answer is 50 W

Hope it helps

8 0
3 years ago
What is the safest way to enter a body of water? Enter with you feet first. Dive in.
zimovet [89]

Answer: feet first

Explanation:

It is always the safest to enter with your feet first. You can become paralyzed or injure you neck if you dive into a body of water that is too shallow. Therefore its always best to dive feet first instead.

4 0
3 years ago
Read 2 more answers
Fill in the blank .Technically color consists of electromagnetic _____
8_murik_8 [283]
Electromagnetic radiation
5 0
3 years ago
Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
A mass of 100g stretches a spring 5cm. If the mass is set in motion from itsequilibrium position with a downward velocity of 10
stepladder [879]

Answer:

The required IVP is;

u'' + 196u = 0

where;

u(0) = 0 and

u'(0) = -10

Explanation:

See the attached for explanation

3 0
3 years ago
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