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Charra [1.4K]
3 years ago
7

Visible light energy.

Physics
2 answers:
Ipatiy [6.2K]3 years ago
6 0
What about it??

Please explain and I will help.

I don’t know what you’re asking.
Leto [7]3 years ago
6 0

Answer:

No se pregunta a alguien mas carnalito.

Explanation:

Lo siento:(

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

7 0
3 years ago
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
3 years ago
A trip is taken that passes through the following points in order
Inessa05 [86]

Answer:

B) - 5.0 m

Explanation:

B is located on a positive location, 15m from the starting point A. Hence, since E is located a positive distance 10m from A, the difference becomes 10 - 15 = - 5.0 m

5 0
3 years ago
determine the maximum angle theta for which the light rays incident on the end of the optical fiber of radius 1 mm are subhect t
Vesna [10]

Answer:

Explanation:

Let the critical angle be C .

sinC = 1 / μ where μ is index of refraction .

sinC = 1 /1.2

= .833

C = 56°

Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )

sin i / sinr = 1.2 , i is angle of incidence

sini = 1.2 x sinr = 1.2 x sin 34 = .67

i = 42°.  

7 0
3 years ago
What is the beat frequency of two keys on a phone, one with a frequency of 210 Hz and the other one with a frequency of 280 Hz?
vodka [1.7K]
The best frequency should be 70 Hz you simply find the difference between 280 Hz and 210 Hz
6 0
3 years ago
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