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xz_007 [3.2K]
3 years ago
15

a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi

ty 0.2, calculate the mass of the fluid in the jar​
Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

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mel-nik [20]

Answer:

at d the charge will be 3q and at 3d it will be 9q

Explanation:

for V=Vp-V2d

V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q

7 0
3 years ago
What is the energy of a 4kg apple that is sitting on a 2 m high tree branch?
miv72 [106K]

78.4 joules is the energy of a 4 kg apple that is sitting on a 2 m high tree branch.

<u>Explanation: </u>

When an apple falls to the ground from a tree, its positional energy (stored as potential gravitational energy) turns into kinetic energy, during a fall. Chemical potential energy is chemical energy because it is food and potential energy as it can still have ability to move. So, in the given case, kinetic energy is zero.

To find potential energy, the formula would be

             \text { potential energy }(P . E)=m \times g \times h

Where, given

m – Mass – 4 kg

g-\text { Earth'sgravity }-9.8 \mathrm{m} / \mathrm{s}^{2} (Known value)

h – Height -  2 m

Substitute these values, we get

                  P . E=4 \times 9.8 \times 2=78.4 \text { joules }

8 0
3 years ago
0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75
UkoKoshka [18]

Answer:

8.0 N

Explanation:

Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).

Mathematically, Fore is expressed as

F = ma ........................... equation 1

Where F = force, m = mass, a = acceleration.

and

I = mΔv

Δv = I/m ............................ Equation 2

Where I = impulse, m = mass, Δv = change in velocity

Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.

Substituting into equation 2

Δv = 6.0/0.1

Δv = 60 m/s.

But

a = Δv/t

where t = time = 0.75 seconds.

a = 60/0.75

a = 80 m/s²

Substitute the values of a and m into equation 1.

F = 0.1(80)

F = 8.0 N.

Thus the average force produced = 8.0 N

6 0
3 years ago
What should you do if a fire breaks out in the back of your boat?
valkas [14]

If a fire breaks out in the back of your boat you should put the back of the boat into the wind so that the fire will not spread easily. Then turn off the engine and paddle the boat to keep the boat into the wind. Also use a fire extinguisher to put the fire off.

8 0
4 years ago
A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. It gets blocked just after re
brilliants [131]
1) Data:

Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m

2) Questions: V at Y = 3m and X at Y = 3 m

3) Calculate components of the initial velocity

Vox = Vo * cos(37°) = 15.97 m/s

Voy = Vo * sin(37°) = 12.04 m/s

4) Formulas

Vx = constant = 15.97 m/s

X = Vx * t

Vy = Voy - g*t

Y = Yo + Voy * t - g (t^2) / 2

5) Calculate t when Y = 3m (first time)

Use g ≈ 9.8 m/s^2

3 = 12.04 * t - 4.9 t^2

=> 4.9 t^2 - 12.04t + 3 = 0

Use the quadratic equation to solve the equation

=> t = 0.28 s and t = 2.18s

First time => t = 0.28 s.

6) Calculate Vy when t = 0.28 s

Vy = 12.04 m/s  - 9.8 * 0.28s = 9.3 m/s

7) Calculate V:

V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s

tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°

Answer: V ≈ 18.5 m/s, with angle ≈ 30°

8) Calculate X at t = 0.28s

X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m

Answer: X ≈ 4,5 m
4 0
3 years ago
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