Answer : The total work done in raising the ball is, 0.98 J
Explanation : Given,
Mass of the ball = 0.075 kg
Height raised of the ball = 1.33 m
As we know that the object is moving with the constant velocity, that means the work done against the gravity will be the net-work done.
So, the work done will be:
where,
w = work done
m = mass of ball
h = height of ball
g = acceleration due of gravity = 
Now put all the given values in the above formula, we get:
Thus, the total work done in raising the ball is, 0.98 J
Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s
Explanation:
Magnetic force(B) = 4.60×10^-3 T
Electric force(E) = 1.64×10^4 V/m
Both forces having equal magnitude ;
Magnetic force = electric force
qvB = qE
vB = E
v = (1.64×10^4) ÷ (4.60×10^-3)
v = 3.57×10^6 m/s
2.) Assume no electric field
qvB = ma
Where a = v^2 ÷ r
R = radius
a = acceleration
v = velocity
qvB = m(v^2 ÷ R)
R = (m×v) ÷ (|q|×B)
q=1.6×10^-19C
m = 9.11×10^-31kg
R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)
R = 32.5227×10^-25 ÷ 7.36×10^-22
R = 4.42×10^-3m
3.) period(T)
T = (2*pi*R) ÷ v
T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)
T = (27.775×10^-3) ÷(3.57×10^6)
T = 7.78×10^-9 s
The resulting positive amplitude of the two waves after the superimposition is 4.30 cm.
<h3>
Amplitude of the waves</h3>
The amplitude of the waves is the maximum displacement of the wave. This is the vertical position of the wave measured from the zero origin.
After the superimposition of the two similar waves, the resulting amplitude will be less than the initial amplitude of the wave with the highest vertical height since the superimposition creates destructive interference.
Resulting amplitude of the two waves is calculated as;
A = 5.4 cm - 1.10 cm
A = 4.30 cm
Thus, the resulting positive amplitude of the two waves after the superimposition is 4.30 cm.
Learn more about amplitude of waves here: brainly.com/question/25699025
<span>(a).the heat trasfer surface area and heat flux on the surface of filament are
Area of Surface= µDL=3.14(0.05cm)(5cm)= 0.785 cm square
qs=Q/Area of surface= 150W/0.785= 191W/cmsq.=1.91x10Âłx10ÂłW/Msq
(b). the heat surface on the surface of heat bulb
Area of surface = 3.14xD²= 3.14(8CM)²= 201.1cm²
qs=Q/Area of surface=150w/201.1cm²=0.75 w/cm²= 7500w/m²
the amount and cost of electrical energy consumed during one period is
Electrical Consumption=QΛt=(0.15 KW)(365X8h/yr)=438 k Wh/yr
Annual cost= 438 kWh/yr)($.08/kWh)= $ 35.04 /yr</span>
It will take 6 hrs since 40•6= 260 I hope this helped
