Answer:
If you were to look for a cut on the palmar surface of a dog's leg then you should look at the back area of the front leg below the carpus.
Explanation:
The value of Q will be -8 C.
In the presence of an electric or magnetic field, matter experiences a force due to its electric charge.
A moving electric charge generates a magnetic field, and an electric charge has an accompanying electric field.
The information provided in the issue is;
The separation between and is 2m.
The separation between and is 2m.
An origin charge equals +2 C
The electric fields are identical in magnitude but are facing in different directions. As a result, the following relationship can be used
Q/16=1/2
The value of Q will be -8 C.
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brainly.com/question/8163163
#4174
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
= 1250W
d) Final peak=
P= Ik/e
= 
P = 2.5 × 10⁷W
Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J
