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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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Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,
otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

  • The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
  • Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
  • If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.

7 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
3 years ago
A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to
dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So 1.41=2\times 3.14\sqrt{\frac{0.683}{K}}

0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

K=9.255N/m

So spring constant of the spring will be equal to 9.255 N /m

7 0
3 years ago
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
4 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
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