What is the momentum of an 80 kg runner moving at the speed of 2.5 m/s? Use the

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 $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago

All isotopes of a particular element have the same atomic number. How then do the isotopes of a particular element differ?

xz_007 [3.2K]

Isotopes of an element will contain the same number of protons and electrons but will differ in the number of neutrons they contain. In other words, isotopes have the same atomic number because they are the same element but have a different atomic mass because they contain a different number of neutrons

7 0

3 years ago

What effect does the sun have on the earth. written response

maw [93]

It had a gravitational force that acts so that it pulls Earth in an orbital, which explains temperature variations throughout the years.

8 0

3 years ago

Suppose each object emitted a burst of light right now.Rank the objects from left to right based on the amount of time itwould t

Wittaler [7]

Answer:

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way > center of the milky way> nevulosa orion> Pluto> Sum

Explanation:

The light that comes from stars and galaxies travels in a vacuum so its speed is constant and with a value of c = 3 108 m / s, so time will be directly proportional to the distance to the object

x = c t

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way > center of the milky way> nevulosa orion> Pluto> Sum

4 0

3 years ago

A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve

kiruha [24]

To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this

Initial Velocity,

$V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$