<u>Answer:</u>
In addition to average weather conditions, climatological data also describes annual variations and fluctuations of temperature, precipitation, wind speed and other variables.
<u>Explanation</u>:
A lot many observations are made all around the world regarding the weather each day. These observations and analysis are done by humans as well as automated instruments. The weather data is collected each day all year and any inaccuracies and discrepancies are checked and rectified. The results are later then presented as the climate data. There are various factors that are taken into consideration while determining the climate of a region. Apart from the factors that are already mentioned, wind speed is also one of the other variables.
Answer:
option (a) 0.61 s
Explanation:
Given;
Time taken by the ball to reach the ground = 0.50 s
Let us first calculate the distance through which the ball falls on the ground
from the Newton's equation of motion, we have

where,
s is the distance
a is the acceleration
t is the time
here it is the case of free fall
thus, a = g = acceleration due to gravity
u = initial speed of the ball = 0
on substituting the values, we get

or
s = 1.225 m
Now,
when the elevator is moving up with speed of 1.0 m/s
the initial speed of the ball = -1.0 m/s (as the elevator is moving in upward direction)
thus , we have

or

or
4.9t^2 - t - 1.225 = 0
or
t = 0.612 s
hence, the correct answer is option (a) 0.61 s
Answer:
331.7m/s
Explanation:
Given parameters:
Initial velocity = 100m/s
Acceleration = 50m/s²
Distance = 1km = 1000m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to apply the right motion equation shown below;
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Now insert the parameters and solve;
v² = 100² + (2 x 50 x 1000)
v² = 110000
v = √110000 = 331.7m/s
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
1.11×10⁸ Electrons.
Explanation:
Applying,
Q = CV..................... Equation 1
Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.
Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V
Substitute this value into equation 1
Q = 1.97×9×10⁻¹²
Q = 1.773×10⁻¹¹ C.
If the charge on one electron = 1.602×10⁻¹⁹ C.
Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹
Number of electron = 1.11×10⁸