Hydrogen is the second most abundant gas in the atmosphere? True or false?

A charging bull elephant with a mass of 5240 kg comes directly toward youwith a speed of 4.55 m/s. You toss a 0.150- kg rubber b

bearhunter [10]

Answer:

Explanation:

mass of elephant, m1 = 5240 kg

mass of ball, m2 = 0.150 kg

initial velocity of elephant, u1 = - 4.55 m/s

initial velocity of ball, u2 = 7.81 m/s

Let the final velocity of ball is v2.

Use the formula of collision

$v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}$

$v_{2}=\left ( \frac{2\times 5240}{5240+0.150} \right )(-4.55)+\left ( \frac{0.15-5240}{5240+0.150} \right )(7.81)$

v2 = - 16.9 m/s

The negative sign shows that the ball bounces back towards you.

(b) It is clear that the velocity of ball increases and hence the kinetic energy of the ball increases. This gain in energy is due to the energy from elephant.

5 0

3 years ago

If a transparent blue filter is placed over a camera lens, what happens to visible light entering it

Leya [2.2K]

Answer:

B

Explanation:

If you pass white light through a blue filter, then only blue color will appear on the other end. This is because a blue filter will only allow blue light through it. And will absorb other color. In other words the rest colors would disappear. Hence the camera would receive just blue light. This phenomenon is referred to as color by subtraction.

5 0

3 years ago

A snowmobile has an initial velocity of 4.7 m/s.

zimovet [89]

Q1. The answer is 8.788 m/s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?

V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s

Q2. The answer is 9.22 s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V2 = 0 (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?

0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s

3 0

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