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3 years ago

In an isolated system, Bicycle 1 and Bicycle 2, each with a mass of 10 kg,

Physics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

momentum before the crash will be:

M1V1 - M2V2

Substitute all the parameters into the formula

10 × 5 - 10 × 3

50 - 30

20 kgm/s

Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s

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2 years ago

B. On a separate sheet of paper, describe the different ways of generating electric power.

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2 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

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3 years ago

1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant

r-ruslan [8.4K]

Answer:

a=0 v = v₀ + a t

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1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

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3 years ago