Answer:
a) 29.36 m
b) 2.44 s
c) 2.57 s
d) 25.117 m/s
Explanation:
t = Time taken
u = Initial velocity = 24 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
b)

Time taken by the ball to reach the highest point is 2.44 seconds
a)

The highest point reached by the ball above its release point is 29.36 m
c) Total height is 3+29.35 = 32.35 m

The ball reaches the ground 2.57 seconds after reaching the highest point
d)

The ball will hit the ground at 25.2117 m/s
Answer:
6 light years = 57 million km
Explanation:
Given;
A light year = 9.5 million km
To calculate how far is 6 light years;
6 light years = 6 × 1 light year = 6 × 9.5 million km
6 light years = 57 million km
Answer:
<em>2 m/s</em>
<em></em>
Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

where
is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
is the velocity of the fluid through the field = ?
is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x 
2.88 x 10^-3 = 1.44 x 10^-3 x 
= 2.88/1.44 = <em>2 m/s</em>
Answer:
The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.
In fact, the magnetic force exerted by the magnetic field on the wire is
where I is the current in the wire, L the length of the wire, B the magnetic field intensity and the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is and so , therefore the magnetic force is zero: F=0.
To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

Here,
= Frequency of Source
= Speed of sound
f = Frequency heard before slowing down
f' = Frequency heard after slowing down
v = Speed of the train before slowing down
So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,
The first equation is,



Now the second expression will be,



Dividing the two expression we have,

Solving for v, we have,

Therefore the speed of the train before and after slowing down is 22.12m/s