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2 years ago

A tide of increased range that occurs during the new and full moons is called

Physics

1 answer:

irina [24]2 years ago

4 0

A tide of increased range that occurs during the new and full moons is called Spring Tide.

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Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

svlad2 [7]

(a) $5.45 \cdot 10^{14} Hz$

The relationship between frequency and wavelength of an electromagnetic wave is given by

$c=f \lambda$

where

$c=3.00 \cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

$\lambda$ is the wavelength

In this problem, we are considering light with wavelength of

$\lambda=5.50 \cdot 10^{-7} m$

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

$f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz$

(b) $1.83\cdot 10^{-15} s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

And using $f=5.45 \cdot 10^{14} Hz$ as we found in the previous part, we can find the period of this wave:

$T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s$

5 0

3 years ago

A coastal forest was home to a type of burrowing rodent that was not found anywhere else in the world. A year ago, a species of

Lapatulllka [165]

Answer:A

Explanation: just took it on E 2020

7 0

3 years ago

A large rock is weathered into tiny pieces which add up the weight of the original rock this demonstrates the conversion of ____

stich3 [128]

When a large rock is weathered into tiny pieces which add up the weight of the original rock, this demonstrates the law of conservation of matter.

According to this law the mass of an object doesn't change with time and also it does not depends on how the particles are arranged themselves.

Hence, option (C) is correct.

5 0

3 years ago

A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta

fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$

4 0

3 years ago

Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

2 years ago