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Aleks04 [339]
3 years ago
6

Which type of waves cannot travel through a vacuum visible light waves x-ray waves gamma ray ray waves or sound waves

Physics
1 answer:
Leni [432]3 years ago
6 0

Answer:

Sound Waves

Explanation:

I remember doing a experiment where there was a bell in a vacuum. When the vacuum was turned on, over time, the bell was sound was drained out.

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Can someone help me?!!!!!
ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

(-3\cdot (-6), -2 \cdot (-6))=(18,12)

The magnitude is given by Pythagorean's theorem:

m=\sqrt{18^2+12^2}=21.6

and the direction is given by the arctan of the ratio between the y-component and the x-component:

\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}

3 0
3 years ago
The alpha line in the balmer series of the hydrogen spectrum consists of light having a wavelength of 6.56. calculate the freque
guajiro [1.7K]
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz. 
5 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
A.Substitute in the units for each one and combine like terms.
storchak [24]
IDK     ghjfnhgfjmrmhjgfhgfmmfh
8 0
3 years ago
A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f
Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

A=\pi r^2

A =\pi (0.2)^2

A =0.1256 m^2

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

\mu=IA

\mu=2(0.1256)

\mu=0.25 A\cdot m^2

Therefore the magnetic dipole moment is 0.25A\cdot m^2

6 0
3 years ago
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