Please need help on this thank you

vA child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0

ikadub [295]

Answer:

$d = 2.26 km$

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

$v_x = 24.8 sin\theta$

$v_y = 24.8 cos\theta$

now the time to reach the child is given as

$\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }$

so now we have

$\theta = 76.5 degree$

So the time to catch the child is given as

$t = \frac{0.6}{24.8 cos78.2}$

$t = 0.104 h$

So distance moved by it in 0.104 h

distance moved by the boat in upstream direction given as

$x = (24.8 sin 74.8 - 3.1)(0.104)$

$x = 2.18 km$

In y direction the displacement of boat is

$y = 0.6 km$

net displacement of the ball is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{2.18^2 + 0.6^2}$

$d = 2.26 km$

4 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$