Answer:
The electric field is 
Explanation:
The force 
on a charge 
in an electric field 
is given by
,
which can be rearranged to give
Now, the force on the electron is 
, and its charge is
; therefore,
Since it is dropped, initial velocity u = 0
Using s = ut + (1/2)gt^2, putting u = 0, g = 10 m/s^2
s =(1/2)gt^2, t = 3s
s = 0.5 * 10 * 3 *3
s = 45 m.
The bridge is 45m above the water.
Answer:
122.735 behind converging lens ; 2.16
Explanation:
Given tgat:
Object distance, u = 29 cm
Image distance, v =
Focal length, f = - 19 (diverging lens)
Mirror formula :
1/u + 1/v = 1/f
1/29 + 1/v = - 1/19
1/v = - 1/19 - 1/29
1/v = −0.087114
v = −11.47916
v = -11.48
Second lens
Object distance :
u = 11.48 + 11 = 22.48 cm
1/v = 1/19 - 1/22.48
1/v = 0.0081475
v = 1 / 0.0081475
v = 122.735 cm
122.735 behind second lens
Magnification, m
m = m1 * m2
m = - v / u
Lens1 :
m1 = -11.48 / 29 = - 0.3958620
m2 = - 122.735 / 22.48 = - 5.4597419
Hence,
- 0.3958620 * - 5.4597419 = 2.16
Answer:
θ = 8.50°
To the nearest angle
θ = 9.0°
the golfer must hit the ball at angle 9° so that it travels 120 feet.
Explanation:
The range of a projectile is the horizontal distance covered by a projectile, which can be written as;
r = (u^2× sin2θ)/g
Where;
r = range
u = initial speed
θ = angle from horizontal
g = acceleration due to gravity
Solving for θ,
sin2θ = rg/u^2
θ = 1/2 × sin⁻¹(rg/u^2) ....1
Given;
r = 120 ft
u = 115 ft/s
g = 9.81m/s = 32.2 ft/s
Substituting the values into the equation 1;
θ = 1/2 × sin⁻¹(120×32.2/115^2)
θ = 1/2 × sin⁻¹(0.29217)
θ = 1/2 × 17.00
θ = 8.50°
To the nearest angle
θ = 9.0°
Answer:
Ws=8.75 Watts
Explanation:
As per fig. of prob 02.061, it is clear that R5 and R4 are in parallel, its equivalent Resistance will be:
Now, this equivalent Req45 is in series with R3, therefore:
This Req345 is in parallel with R2, i.e
Now this gets in series with R1:
Now, the power delivered Ws is:
