Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
Answer:
Explanation:
Using kinematics equations:
Use 
due to condition of distance traveled.
Solving second equation for time, there are two solutions. t=0 and
Use the expression in the first equation to have
Using trigonometric identities, you have the answer of the distance.
By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.
Answer: The correct explanation is 2.
Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.
On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.
Answer:
option D
Explanation:
given,
A conductor is carrying current = 2.0 A is 0.5 mm thick
Hall voltage = 4.5 x 10-6 V
uniform magnetic field = 1.2 T
density of the charge = n =?
hall voltage =
n = 6.67 × 10²⁷ charges/m
hence the correct answer is option D
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
