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juin [17]
4 years ago
8

A boat has a mass of 31,999 kg and a plane has a mass of 110,000 kg. What is the ratio of weight of the plane to the boat?

Physics
2 answers:
Gelneren [198K]4 years ago
8 0

Answer:

3.44:1

Explanation:

We use ratios to find the ratio of weight.

 Mass of Plane= 110000 kg                          ∴ W= mg

 Mass of  Boat =31999 kg

\frac{weight of plane}{weight of boat}=\frac{110000*9.81}{31999*9.81}

 =\frac{3.44}{1}

SO 3.44:1

Helga [31]4 years ago
6 0
<h2>Answer:</h2>

A. 3.44 :1

<h2>Explanation:</h2>

We use ratios to compare things, that is, this helps us to know <em>how much of one thing there is compared to another thing. </em>We can express ratios in different ways:

Using \ fractions: \ \frac{a}{b} \\ \\ \\ Using \ words: \ a \ to \ b \\ \\ \\ Using \ : \ to \ separate \ values

In this exercise, we know that:

  • A boat has a mass of 31,999 kg
  • A plane has a mass of 110,000 kg

And we want to know the ratio of weight of the plane to the boat. So, if:

a: mass of the boat 31,999 kg

b: mass of the plane 110,000 kg

Then, the ratio of weight of the plane to the boat is:

\frac{b}{a}=\frac{110,000}{31,999}=3.44 \\ \\ \\ So: \\ \\  \boxed{\frac{b}{a}=3.44:1}

<em>So the correct option is A. 3.44:1</em>

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From the graph attached,

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Describe what happens when you jump from a small boat onto a dock from the perspective of the 3rd Law.
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5 0
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A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s
ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = mu + 0 = mu

Momentum just after collision = (m + M)v

Therefore, mu=(M+m)v --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

f=\mu mg=-ma\\\\a=-\mu g

Now, using equation of motion, we can find the velocity just after collision.

0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}

Plug in the given values and find 'v'. This gives,

v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s

Now, using equation (1) and substituting the given values, we get:

0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

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