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3 years ago

A box has a length of 18.1 cm and a width of 19.2cm tall ?

Physics

1 answer:

ladessa [460]3 years ago

3 0

Volume = l*w*h = (18.1 cm)(19.2 cm)(20.3 cm) = 7,054 cm^3.

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Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85

vichka [17]

The force require to keep grouper submerged is 8.207N.

According to Archimedes principle buoyant force of any object must equal to weight of fluid it displaced.

The expression for the force exerted to stay submerged in salt water is

F = F(b) - w(fish)

where F(b) = buoyant force

w(fish) = weight

now substitute w(b) for F(b)

→ F = Vρg - w(fish)

where V = volume of sea water

ρ = density of sea water

Now by Archimedes principle V = m(fish) / ρ(fish)

→ F = (m(fish) / ρ (fish) ) ρg - m(fish)g

F = (85 kg/1015 kg-m^-3) (1.025× 10³ kg-m^-3) (9.8 m/s^2)

- (85kg) × 9.8 m/s^2

F = 841.207N - 833N

F = 8.207 N

Hence, the force require to keep grouper submerged is 8.207N.

Learn more about Archimedes Principle here:

brainly.com/question/15076878

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7 0

2 years ago

Planck's constant, h, is 6.626 x 10-34 js. the speed of light in a vacuum, c, is 3.00 x 108 m/s. calculate the frequency of a gr

Misha Larkins [42]

Frequency= velocity of light/wave length
Fr= 3×10^8/510×10^-9
Frequwency=5.88×10^14 Hz

4 0

3 years ago

Explain why you can "stack" water slightly above the rim of a glass if you pour the water in very carefully?

jeka57 [31]

I believe Isaac Newtons Law " Gravity ".

8 0

3 years ago

A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f

Ganezh [65]

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

8 0

3 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

3 0

3 years ago