Answer:
m = 4
Explanation:
We have,
You apply a force of 600 N to the branch which acts as a lever. It means it is input force, IF = 600 N
The rear of the truck weighs 2,400 N. It means it is output force, OF = 2400 N
The ratio of output force to the input force is equal to the mechanical advantage of the lever arm. It is given by :

So, the mechanical gain of the lever arm is 4.
Answer:
a) FE = 0.764FG
b) a = 2.30 m/s^2
Explanation:
a) To compare the gravitational and electric force over the particle you calculate the following ratio:
(1)
FE: electric force
FG: gravitational force
q: charge of the particle = 1.6*10^-19 C
g: gravitational acceleration = 9.8 m/s^2
E: electric field = 103N/C
m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg
You replace the values of all parameters in the equation (1):

Then, the gravitational force is 0.764 times the electric force on the particle
b)
The acceleration of the particle is obtained by using the second Newton law:

you replace the values of all variables:

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.
Answer:
The store energy in the inductor is 0.088 J
Explanation:
Given that,
Inductor = 100 mH
Resistance = 6.0 Ω
Voltage = 12 V
Internal resistance = 3.0 Ω
We need to calculate the current
Using ohm's law


Put the value into the formula


We need to calculate the store energy in the inductor



Hence, The store energy in the inductor is 0.088 J
<u>First Symbol </u>: Cobalt (Co)
Its Group Number - 9
Its Period Number - 4
Its Family Name - Transition Metal
<u>Second Symbol</u> : Silicon (Si)
Its Group Number - 14
Its Period Number - 2
Its Family Name - Semiconductor
<u>Third Symbol</u> : Astatine (At)
Its Group Number - 17
Its Period Number - 6
Its Family Name - Halogen
<u>Fourth Symbol </u>: Magnesium (Mg)
Its Group Number - 2
Its Period Number - 3
Its Family Name - Alkaline Earth Metal
<u>Fifth Symbol</u> : Xenon (Xe)
Its Group Number - 18
Its Period Number - 5
Its Family Name - Noble Gas
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x
m/s)
f= 3 x
/ 2.4
f=1.25 x
hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x
=> 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m