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Leviafan [203]
4 years ago
12

What wavelength of light would you report in units of nm, if the light had a wavelength of 7.60 x 10-10 m?

Physics
1 answer:
mezya [45]4 years ago
4 0

Explanation:

One meter is equal to 10^{9}nm, so:

7.60*10^{-10}m*\frac{10^9nm}{1m}=0.76nm

This wavelength does not correspond to any color of the visible spectrum. A wavelength of 0.76nm would correspond to soft X-rays, since these electromagnetic radiation have a wavelegth range from 1nm to 0.1nm.

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Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang
Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

                                         ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

                              2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0
3 years ago
How does the direction of acceleration compare with the direction of the net force that produces it
Novay_Z [31]

Answer:

Explanation:

The direction of the acceleration is in the same direction as the net force causing it.  F = ma is actually a vector equation in which f and a are both vectors and m is a scalar constant.

8 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
List two industrial uses for nonmetallic mineral resources
Anni [7]

Answer:

To make useful chemicals and abrasives

Explanation:

4 0
3 years ago
A sled is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy isJ. (Formula: PE =
elixir [45]
Data:

h = 2m
m = 45 Kg
PE = ? (Joule)

Adopting, gravity (g) ≈ <span>9,8 m/s² 
</span>
Formula: PE_{grav}  = mass * g * height

Solving:

PE_{grav} = mass * g * height
PE_{grav} = 45*9,8*2
PE_{grav} = 882J

Answer:
<span>The sled's potential energy is 882 Joules</span>
8 0
3 years ago
Read 2 more answers
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