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vfiekz [6]
1 year ago
9

A petrol tanker ha 2800kg when empty and hold 30m3 0f petrol when full. The denity of petrol i 740kg/m3. Calculate the ma of the

tanker when full of petrol
Physics
1 answer:
White raven [17]1 year ago
4 0

The mass of the tanker with petrol is 250000 N.

We are given that,

Mass of tanker= m =2800 kg

Volume of petrol= v =30 m³

Density of petrol= d =740 kg/m³

Thus , mass , density and volume relation can be given as,

density= mass/ Volume

Mass = Density × Volume

Mass = 740× 30

Mass = 22200 kg

The mass of the petrol is 22200 kg.

Total mass of tanker with petrol = Mass of petrol + Mass of tanker

Total mass of tanker with petrol= 22200+ 2800 kg

Total mass of tanker with petrol= 25000 kg

Total weight of the tanker with petrol = Mass of tanker full of petrol× g

Where, weight = m × g ,(g =10m/s²)

Total weight of the tanker with petrol= 25000×10 = 250000 N

Therefore, the mass of petrol, total mass of tanker with petrol and weight of tanker with petrol would be  22200 kg, 25000 kg and 250000 N.

To know more about mass

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Explanation:

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Microwave transmission involves sending signals from one microwave station to another. What is this often called?
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3 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

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3 years ago
An old-fashioned LP record rotates at 33 1/3 RPM
Ksenya-84 [330]

Answer:

Part a) \frac{5}{9}\ \frac{rev}{sec}

Part b) \frac{9}{5}\ \frac{sec}{rev}

Explanation:

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we have that

An old-fashioned LP record rotates at 33 1/3 RPM

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33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}

Simplify

\frac{5}{9}\ \frac{rev}{sec}

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3 years ago
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
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Answer:

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given,

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Using equation to solve the problem

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