Answer:
Explanation:
We define the linear density of charge as:
Where L is the rod's length, in this case the semicircle's length L = πr
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dq to center of the circle
Thus.
Potential at the center of the semicircle
Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Answer:
a) 0.0625 I_1
b) 3.16 m
Explanation:
<u>Concepts and Principles </u>
The intensity at a distance r from a point source that emits waves of power P is given as:
I=P/4π*r^2 (1)
<u>Given Data</u>
f (frequency of the tuning fork) = 250 Hz
I_1 is the intensity at the source a distance r_1 = I m from the source.
<u>Required Data</u>
- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.
- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.
<u>solution:</u>
(a)
According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:
I∝1/r^2
Set the proportionality:
I_1/I_2=(r_2/r_1)^2 (2)
Solve for I_2 :
I_2=I_1(r_2/r_1)^2
I_2=0.0625 I_1
(b)
Solve Equation (2) for r_2:
r_2=(√I_1/I_2)*r_1
where I_2 = (1/10)*I_1:
r_2=(√I_1/1/10*I_1)*r_1
=3.16 m
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
