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Anika [276]
3 years ago
9

The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden bl

ock is submerged. Is the unknown liquid more or less dense than water?
Physics
1 answer:
anygoal [31]3 years ago
4 0

Answer:

1.5 g/cm^3

Explanation:

In the question i am assuming that wooden block of weight W floats with exactly one-half of its volume below the waterline.

Let ρ_{liquid} be the density of the unknown liquid , V= volume submerged

g= acceleration due to gravity

then

weight of block = \rho_{water}\times V\times g/2

when the block is in water

Also,

weight of block = \rho_{liquid}\times V\times g/3

dividing eq 1 by eq 2 we get

\rho_{liquid} = 3/2 \rho_{water}

⇒\rho_{liquid} = 3/2

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A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s
abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved.  Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

8 0
3 years ago
Why is electrical energy considered as potential energy as its the energy of moving electrons?
Veseljchak [2.6K]

Answer:

Explanation: When the electrons move in another direction, they convert this chemical potential energy to electricity in the circuit, thus discharging the battery. So, the battery is all potential energy.

8 0
2 years ago
Read 2 more answers
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
What resistance would produce a current of 120A from a 6-volt battery?
andriy [413]
120/6=20 ohms :) would be the ans
6 0
3 years ago
The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between
Nat2105 [25]

Answer:

14 x 0.27 = 3.78 is your answer

Explanation:

the question is asking for the weight of the object so you multiply and get 3.79

5 0
3 years ago
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