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marissa [1.9K]
3 years ago
12

A submarine emerges 1/9 of its volume when it partially floats on the sea surface. For make it completely submerge it is necessa

ry to allow a volume of 50,000 [L] to enter of sea water whose density is 1,026 [g / cm3]. What is the weight of the submarine?
Physics
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

4,524,660 N

Explanation:

Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.

m/9 = (1026 kg/m³) (50 m³)

m = 461,700 kg

mg = 4,524,660 N

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?​
NeX [460]
Hey there!

We are given ,

Acceleration, a = -2m/s^2

Initial velocity , u = 15m/s

Time , t = 5 seconds

We know that ,
V=u+at

Now , final speed ,

V = 15+(-2)(5)

V = 15-10

V = 5 m/s -> final speed

Hope this helps you dear :)
Have a good day <3

8 0
2 years ago
Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat
jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{24}

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While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{72}

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Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0
3 years ago
Sheila eats an energy bar right before she goes on a run. What energy changes will occur in sheila as she runs after eating?
LiRa [457]
The energy bar eaten by Sheila has chemical energy locked up inside it. This chemical energy is converted to mechanical energy in form of potential and kinetic energy and this in turn is converted to heat energy as the run progresses. Thus, the energy changes are: chemical energy to mechanical energy [kinetic and potential] and finally to heat energy.
8 0
3 years ago
Read 2 more answers
Mechanical energy is a term that is used to describe
larisa86 [58]
The sum of potential energy<span> and kinetic </span><span>energy.
Hope I helped!</span>
7 0
4 years ago
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