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Vsevolod [243]
3 years ago
10

What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

in angstroms (x 10-10 m) to 2 sig figs.
Physics
1 answer:
ad-work [718]3 years ago
6 0

Answer:

1.2826 x 10^-13 m

Explanation:

\lambda  = \frac{h}{\sqrt{2 m K}}

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}

λ = 1.2826 x 10^-13 m

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
How did water get from the ocean to your water faucet?
Semmy [17]

Answer:

lol im pretty sure pipes and nice pic of lil darkie

Explanation:

2+2=4

3 0
3 years ago
Read 2 more answers
What is the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head? The radius of the cir
Leya [2.2K]
Well the centripetal acceleration would be 9.82 squared. That is the correct answer. I hope this helped you. Have a great day! :)
4 0
3 years ago
Read 2 more answers
If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?
never [62]

The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

Explanation:

When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.

spring constant is k = 1.00 * 10^3 N/m

mass is x = 20.0 cm

According to Hooke's law, To find restoring force,

    F = - kx

        = - 1.00 *10 ^3 * 20.0

     F = 20000 N/m

Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

3 0
3 years ago
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