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Bond [772]
2 years ago
14

What is the wavelength of a wave that has a frequency of 15HZ and a wave speed of 5 m/s?

Physics
1 answer:
stepladder [879]2 years ago
5 0

Answer:

A.   .33m

Explanation:

Have a good day.

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Give me 1 example of complete inelastic collision​
Allisa [31]

i hope it helped thanks

4 0
3 years ago
A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc
blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0
3 years ago
Read 2 more answers
All of the planets and their satellites orbit the Sun in the same direction, and all their orbits lie near the same plane. True
MissTica

Answer:

True

Explanation:

The Sun rotates in the counterclockwise (CCW) direction when seen from its north pole. Since, the planets revolve around the Sun because of its gravity, the revolution of all the planets and their moons as seen from the north of the Sun is in CCW direction.

In fact most of the solar system bodies rotate in the same direction that is CCW. Some major exceptions to this are Venus and Uranus.

Almost all the planets and moons were made from the planetary disk around the Sun. Thus, they lie nearly in the same plane.

4 0
3 years ago
Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0
3 years ago
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

4 0
3 years ago
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