Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
P = 180 [w]
Explanation:
To solve this problem we must use ohm's law, which is defined by the following formula.
V = I*R & P = V*I
where:
V = voltage = 200[volts]
I = current [amp]
R = resistance [ohm]
P = power [watts]
Since the bulbs are connected in series, the powers should be summed
P = 60 + 60 + 60
P = 180 [watts]
Now we can calculate the current
I = 180/200
I = 0.9[amp]
Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.
And the power is defined by P = V*I
we know that the voltage is equal to 200[V], therefore
P = 200*0.9
P = 180 [w]
Answer:
The time taken by the projectile to hit the ground is 6.85 sec.
Explanation:
Given that,
Vertical height of cliff = 230 m
Distance = 300 m
Suppose, determine the time taken by the projectile to hit the ground.
We need to calculate the time
Using second equation of motion

Where, s = vertical height of cliff
u = initial vertical velocity
g = acceleration due to gravity
Put the value in the equation



Hence, The time taken by the projectile to hit the ground is 6.85 sec.
Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters