Answer: P= mad/t or P=w/t so P= 300/6= 50 W
Answer:
Often loose fibers surrounding a contact hole show scorch marks from the flame discharge of the weapon. Some synthetic fibers may be melted. The blowback of muzzle gases may produce a star shaped tear pattern around the hole.
Explanation:
Answer:
(a). = 5.43
(b). 13
(C) 0
Explanation:
part A:
φ(outside) = ∬B(outside) dS
note that;
φ(sum) = 2[φ(outside) + φ(inside).
then, we say
φ(outside) = ∫ μI/ 2πr dr (taking boundaries at R - d/2 and d/2.
μI/2πr dr= μI/2πr ln 2R- d/d
= 1.26× 1) ^-6 ×10 ln 40-2.5/2.5.
= 5.43 μWb.
magnetic flux in the conductor can be calculated from magnetic induction as integral of d/2 to R
please note that dS= 1. dr
φ (inside) = B(inside) dS
∫2μIr/ π d^2 dV.
μI/πd2 r^2 (at boundary d/2 and 0)
μI/4π
= 1.26×10^-6 ×10/4π.
= 1.0032 μWb .
where B(inside) = μI(inside)/ 2πr and I= Ir^2/ (d/2) ^2.
φ(sum)= φ(outside + inside)
=2(5.42 + 1.0032
= 13μWb
(c). is zero
Answer:
8.436 Nm
16.872 Nm
0
Explanation:
When we multiply the force with the distance at which the force is being applied we get torque
where
F = Force = 111 N
r = Distance = 0.152 m
= Angle at which force is applied
When,
The torque is 8.436 Nm
When,
The torque is 16.872 Nm
When,
The torque is 0
Answer:
V=120m/s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem you must divide the problem into two parts 1 and 2, when the rocket accelerates (1), and when the rocket decelerates (2).
Then you raise equations 3 and 1 in both parts.
finally you use algebraic methods to find the value of speed
I attach the complete procedure