Answer:
36.87°
Explanation:
Given
Maximum height = 60m
Horizontal distance (range) = 80,m
Required
Initial speed U
Angle of launch
To get the speed, we will use the range formula;
R = U √2H/g
80 = U√2(60)/9.8
80 = U√12.25
80 = 3.5U
U = 80/3.5
U = 22.86m/s
Get the angle of launch
Using the formula
Theta = tan^-1(y/x)
y is the vertical distance
x is the horizontal distance
Theta = tan^-1(60/80)
Theta = tan^-1(0.75)
Theta = 36.87°
Hence the angle of launch is 36.87°
Answer:
B. 2m/s² south
Explanation:
Given parameters:
Net force = 40N
Mass = 20kg
Unknown:
Acceleration of the object = ?
Solution:
To solve this problem, we use;
F = ma
F is the net force
m is the mass
a is the acceleration
a =
a = = 2m/s²
The acceleration of the object is 2m/s² south
It's not. It's equivalent to the transfer of that many elementary charges <u>every second</u> .
The elementary charge (charge on an electron) is
1.60 x 10⁻¹⁹ coulomb / electron.
The number of electrons in a Coulomb is the reciprocal of that number.
6.25 x 10⁻¹⁸ electrons / Coulomb.
There's the number. 1 Coulomb <em>per second</em> is called 1 Ampere.
Answer:
Current amplitude, I = 5.57 A
Explanation:
It is given that,
Effective resistance, R = 22 ohms
Inductive reactance, L = 72 ohms
Voltage of the alternating source, V = 420 volt
The total impedance of the RL circuit is given by :
Z = 75.28 ohms
From Ohm's law,
I = 5.57 A
So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.
The suns gravity would effect the earth's orbit and the earth would slip out of its normal orbit. Hope this helps.