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Alisiya [41]
4 years ago
14

Based on the information contained in the figure above, do you think that neon has any isotopes?

Physics
1 answer:
Yakvenalex [24]4 years ago
6 0
Yes neon does have isotopes
You might be interested in
I can’t figure this out!!! Answer what you can , please.
tigry1 [53]

a/b. The ball has velocity vector at time t

\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)

where v_0=16\dfrac{\rm m}{\rm s} is the ball's initial speed and g=9.8\dfrac{\rm m}{\mathrm s^2}.

c. At its highest point, the ball has 0 vertical speed. This occurs when

v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s

d. Recall that

{v_y}^2-{v_{0y}}^2=-2g\Delta y

so that at its highest point,

0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m

e. This is just twice the time it takes for the ball to reach its maximum height, t=2.9\,\mathrm s.

f. The ball's horizontal position after time t is

v_0\cos63^\circ\,t

so that after the time found in part (f), the ball has traveled

v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m

4 0
4 years ago
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b > a, is filled with
AURORKA [14]

Answer:

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

Explanation:

So, we will be making use of the data or parameters given in the question above;

=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."

=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."

So, we will be making use of the two equations below;

Ë = ( λ/ 2πEP) × P'. --------------------------(1)..

B' = √ μE × ( λ/ 2πEP) × P' --------------(2).

Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.

Jo = 1/√ μE × λ/2 × π × a.

When we solve for charge per unit length, we have;

λ = 2 × π × Jo × a × √ μE.

The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.

Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.

Therefore, P = ∫z' . <s> da

P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.

(Take limit on the first at second integration as : 2π,0 and b,a).

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

5 0
3 years ago
If a machine exerts a force of 250 N on an object and no work is done, what must have occurred?
valina [46]

Answer:

<h2>1) There is no work done by the machine because</h2><h2>B) The object has not moved</h2><h2>2) There is no work done by the prisoner because</h2><h2>D) The prisoner does no work because the wall goes no distance</h2><h2>3) The kinetic energy when it is half the way down is</h2><h2>6.0 J</h2>

Explanation:

1) As we know that the work done is the product of force and displacement

It is given as

W = Fdcos\theta

so if the object is not displaced due to the force exerted by the object then the work done by the object must be ZERO

so correct answer is

B) The object has not moved

2) As we know that the work done is the product of force and displacement

It is given as

W = Fdcos\theta

As we know that the wall is not displaced due to applied force so here work done by the prisoner must be zero

D) The prisoner does no work because the wall goes no distance

3) As we know by work energy theorem that work done by all forces is equal to change in its kinetic energy

So we will have

W_g + W_f = \frac{1}{2} mv^2

so we will have

3(10)(4) + W_f = \frac{1}{2}(3)(3)^2

120 + W_f = 13.5

W_f = -106.5 J

now when cart moves half the distance then again using the same

W_g + W_f = K

K = 3(10)(2) - \frac{106.5}{2}

K = 6.5 J

8 0
3 years ago
A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t
Mama L [17]

Answer:

c. √2T

Explanation:

The period of a simple pendulum is given by;

T = 2\pi \sqrt{\frac{L}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}} \\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\\frac{g}{4\pi^2}  = \frac{L}{T^2}\\\\ \frac{L_1}{T_1^2}= \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{L_2T_1^2}{L_1}\\\\L_2 = 2L_1\\\\ T_2^2 = \frac{2L_1T_1^2}{L_1}\\\\ T_2^2 =2T_1^2\\\\T_2 = \sqrt{2T_1^2}\\\\T_2 = T_1\sqrt{2}

Thus, the the new period will be √2T

4 0
3 years ago
An object moving at 30 m/s takes 5s to come to a stop. what is the object's acceleration?
Taya2010 [7]
Use SUVAT where:
S = distance
U = initial velocity = 30m/s
V = end velocity = 0 m/s
A = acceleration = ?
T = time = 5 secs

then pick out the equation v = u + at

so:
0 = 30 + 5a
-30 / 5 = a
a = -6 m/s^2
4 0
3 years ago
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