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WITCHER [35]
2 years ago
8

A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. show answer No Attempt 50%

Part (a) If the canoeist gets up and walks to the front of the canoe and there is no friction between the boat and the water, how far will they be from the dock
Physics
1 answer:
Pavel [41]2 years ago
6 0

The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

v(m₁ + m₂) = m₁v₁ + m₂v₂

where;

v is the velocity of the canoeist and the canoe when they are together

  • u₁ is the velocity of the canoe
  • u₂ velocity of the canoeist
  • m₁ mass of the canoe
  • m₂ mass of the canoeist

<h3>Distance traveled by the canoeist</h3>

The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.

Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?
Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0
3 years ago
Add the vectors:
Anettt [7]

Vector 1 has components

x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m

y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m

and vector 2 has

x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m

y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m

Add these vectors to get the resultant, which has components

x_{\rm total}\approx11.133\,\mathrm m

y_{\rm total}\approx13.268\,\mathrm m

The magnitude of the resultant is

\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m

with direction \theta such that

\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ

or about 50º N of E.

8 0
3 years ago
What is the formula to calculate force?
Juliette [100K]
The process of determining the value of the individual forces acting upon an object involve an application of Newton's second law (Fnet<span>=m. a) and an application of the meaning of the </span>net force<span>.</span>
3 0
4 years ago
Three identical very dense masses of 3500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,
larisa86 [58]

Answer:

<em>A) 7.37 x 10^-4 N</em>

<em>B) The resultant force will be towards the -x axis</em>

Explanation:

The three masses have mass = 3500 kg

For the force of attraction between the mass at the origin and the mass -100 cm away:

distance r = 100 cm = 1 m

gravitational constant G= 6.67×10^−11 N⋅m^2/kg^2

Gravitational force of attraction F_{g} = \frac{Gm^{2} }{r^{2} }

where G is the gravitational constant

m is the mass of each of the masses

r is the distance apart = 1 m

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{1^{2} } = 8.17 x 10^-4 N

For the force of attraction between the mass at the origin and the mass 320 cm away

distance r = 320 cm = 3.2 m

F_{g} = \frac{Gm^{2} }{r^{2} }

substituting, we have

F_{g} = \frac{6.67*10^{-11}*3500^{2} }{3.2^{2} } = 7.98 x 10^-5 N

Resultant force = (8.17 x 10^-4 N) - (7.98 x 10^-5 N) = <em>7.37 x 10^-4 N</em>

<em></em>

<em>B) The resultant force will be towards the -x axis</em>

7 0
3 years ago
ASAP URGENT A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in
Umnica [9.8K]

Answer:

8018.2 N

Explanation:

Change in momentum, p is equivalent to force * time

M(vf-vi)=Ft where M represent mass of the car, vf is the final speed, vi is the intial speed of the car, F is the force in queation and t ia the time.

Since the car finally comes to rest then the final velocity ia zero. Substituting 1400 Kg for M, 0 for vf, 27 m/s for vi and 1.5 s for t then the first force is

1400(0-27)=1.5F

F=25200 N

Similarly, when we work with the same values but t changed to 2.2s

1400(0-27)=2.2 F

F2=17181.818181818 N

Reduced force will be 25200-17181.818181818=−8,018.181818182N

Approximately, 8018.2 N

3 0
3 years ago
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