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4 years ago

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If an electric train is moving north at 100mph and the wind is blowing to the west at 10mph, which way does the smoke blow?

1 answer:

Yakvenalex [24]4 years ago

8 0

If an electric train is moving north at 100mph and the wind is blowing to the west at 10mph, which way does the smoke blow?

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combining the velocity vectors

angle whose tangent is 10/100 = 0.1

angle whose tan is 0.1 north west

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While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

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While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

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V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

<span>(b) </span>The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

<span>(c) </span>If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

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4 years ago

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An object carrying a force of 15N has a mass of 3kg. What is the<br> acceleration of this object?

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Calculate the average power transmitted by the string. data: mass of a 147 m long piece of the string is 2.19 kg.

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1 year ago

Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

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3 years ago