Answer:
75joules
Explanation:
Workdone = force x distance
workdone = 25 x 3
workdone = 75joules
Because we can reproduce with just one and if more eggs were released, the female reproductive life (which is 40 years) would be shorter and so there would be less time for every female human to reproduce since the amount of eggs is limited. (Also fraternal twins come from 2 eggs released at the same time)
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:
<span>centripetal accel = 1.5*g
ω²r = 1.5*9.8m/s²
ω² * 8m = 14.7 m/s²
ω = 1.36 rad/s * 1rev/2πrads * 60s/min = 12.9 rpm</span>
Answer:
3.528×10² V.
Explanation:
potential difference: This is the work done when one coulomb of charge moves from one point to another in an electric field. The S.I unit of potential difference is volt. The formula of potential difference is given as,
V = kq/r..................... Equation 1
And
E = kq/r² .................. Equation 2
Comparing equation 1 and equation 2,
V = E×r............................. Equation 3
Where V = potential difference, E = Electric field between the plate of the capacitor, r = distance between the plate.
Given: E = 6.3×10⁵ V/m, r = 0.56 mm = 0.00056 m.
Substitute into equation 3,
V = 6.3×10⁵×0.00056
V = 3.528×10² V.
Hence the potential difference of the plate = 3.528×10² V.
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
