Answer:
E = 31.329 N/C.
Explanation:
The differential electric field 
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where 
is the radius of curvature.
Now
,
where 
is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find 
and 
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
Answer:
4 gamma closest thing to this V
Explanation:
Technetium. Tc is a very versatile radioisotope, and is the most commonly used radioisotope tracer in medicine.
The answer is c.
Sound, light and heat energy.
Hope this helped :)
Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
Answer:
D) 15s
Explanation:
let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.
the period of a pendulum is given by:
T = 2π√(L/g)
so on earth:
Te = 2π√(L/g1)
= 6s
on the moon;
Tm = 2π√(L/g2)
since g2 = 1/6 g1 then:
Tm = 2π√(L/(1/6×g1))
= √(6)×2π√(L/(g1))
and 2π√(L/(g1)) = Te = 6s
Tm = (√(6))×6 = 14.7s ≈ 15s
Therefore, the period of the block-spring system on the moon is 15s.
