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2 years ago

What is the wave speed of a wave with a period of 5 seconds and a wavelength of 50 meters?

Physics

1 answer:

devlian [24]2 years ago

8 0

Answer:

10m/s

Explanation:

speed is calculated by taking distance covered then divided by time taken .

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What is the velocity in meters per second of a runner who runs exactly 110 m toward the beah in 72 seconds?

rjkz [21]

B) 1.53 m/s towards the beach

8 0

3 years ago

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Who was the man who lived from 460B.C. 370B.C. and was among the first to suggest the idea of atoms?

Jlenok [28]

C Democritus

That's the Answer

6 0

3 years ago

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When a planet's orbit takes it closest to the Sun, its called:________.

andriy [413]

Answer:

perihelion

Explanation:

The point at which a planet is closest to the sun is called perihelion. The farthest point is called aphelion

3 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

3 years ago

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Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

4 years ago