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4 years ago

Which of the following is of particular interest to scientists in their observations because of potential threats to the Earth?

Physics

1 answer:

nordsb [41]4 years ago

8 0

The answer is Near Earth Objects (NEOs).

Near Earth Objects (NEOs) have gained particular interest to scientists in their observations because of its potential threats to the Earth. It can be comets or asteroids. It is called Near Earth since their orbits are too close to cross the orbit of the Earth thus NEO can reach the space near-Earth that more likely give potential threats to the Earth.

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Through which of these media do sound waves

defon

Wood i think could be wrong

3 0

3 years ago

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.50 m .How many times must he repeat this exercise to burn

ludmilkaskok [199]

Answer:

$N = 2141 times$

Explanation:

Each time the work done to raise a given mass is

$W = mgh$

here we know that

$m = 30 kg$

$h = 0.50 m$

now we have

$W = (30 kg)(9.81 m/s^2)(0.50)$

$W = 147.15 J$

since it is just 25% of actual energy consumed as we know its efficiency is 25%

so we have total energy consumed in this way

$E_{total} = \frac{147.15}{0.25}$

$E_{total} = 588.6 J$

now if it took N number of times so burn the fat of a pizza then

$N(588.6) = 1260 \times 10^3$

$N = 2141 times$

4 0

3 years ago

PLEASE HELPPPP MEEE!!!!!!!!!!!!!!!

ozzi

The answer is choice C.

6 0

3 years ago

Read 2 more answers

The wavelength of red helium-neon laser light in air is 632.8 nm.

e-lub [12.9K]

(a) $4.74\cdot 10^14 Hz$

The frequency of an electromagnetic wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of the wave in a vacuum (speed of light)

$\lambda$ is the wavelength

In this problem, we have laser light with wavelength

$\lambda=632.8 nm=6.33\cdot 10^{-7} m$ . Substituting into the formula, we find its frequency:

$f=\frac{3.0\cdot 10^8 m/s}{6.33\cdot 10^{-7} m}=4.74\cdot 10^14 Hz$

(b) 427.6 nm

The wavelength of an electromagnetic wave in a medium is given by:

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the original wavelength in a vacuum (approximately equal to that in air)

$n$ is the index of refraction of the medium

In this problem, we have

$\lambda_0=632.8 nm$

n = 1.48 (index of refraction of glass)

Substituting into the formula,

$\lambda=\frac{632.8 nm}{1.48}=427.6 nm$

(c) $2.03\cdot 10^8 m/s$

The speed of an electromagnetic wave in a medium is

$v=\frac{c}{n}$

where c is the speed of light in a vacuum and n is the refractive index of the medium.

Since in this problem n=1.48, we find

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.03\cdot 10^8 m/s$

3 0

4 years ago

The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a

Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0

3 years ago