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DochEvi [55]
3 years ago
12

In order for a ship to stay afloat, its buoyant force must be

Physics
1 answer:
Kruka [31]3 years ago
4 0
The buoyant force must be greater than water.
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The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to
LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power  ,P = F .v

F=force

v=velocity

v= 100 km/h

v=\dfrac{1000}{3600}\times 100\ m/s

v=27.77 m/s

75 x 1000  = F x 27.77

F= \dfrac{75000}{27.77}\ N

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

 a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0
3 years ago
Where is visible light found on the electromagnetic spectrum?
vampirchik [111]

Answer:

C: in between infrared and ultraviolet light

Explanation:

the colors are

V - violet

I - indigo

B - blue

G - green

Y - yellow

O - orange

R - red

5 0
3 years ago
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl
finlep [7]

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )

The buoyant force of lead block;

F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173  \ N

The buoyant force of copper block

F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219  \ N

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

3 0
3 years ago
Anther graph question pls help thank you
Lera25 [3.4K]
I believe the answer to be she stopped for 4 mins
8 0
2 years ago
Read 2 more answers
When the ball of the pendulum moves from (x) to (y) in a duration of 0.02 sec the frequency equals
Mariulka [41]
0.4 sec the frequency equals
5 0
1 year ago
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