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4 years ago

11

Which type(s) of electromagnetic radiation emitted by the Sun are absorbed by Earth’s atmosphere and do not reach Earth’s surfac

e?
A. long-wave radio waves, X-rays, and gamma rays
B. short-wave radio waves and gamma rays
C. infrared light, X-rays, and gamma rays
D. X-rays

1 answer:

gregori [183]4 years ago

5 0

Gamma rays, X-rays, most ultraviolet rays, and some infrared are absorbed by the atmosphere but do not reach the Earth's surface

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A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0

3 years ago

What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10

raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

$Y = \frac{FL}{A\Delta L}$

$20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}$

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

7 0

4 years ago

Which type of power plants generate the least amount of pollution? A) Biomass power plants B) Geothermal power plants C) Nuclear

sladkih [1.3K]

Answer:

C

Explanation:

6 0

3 years ago

A 20 kg girl runs up the staircase to a floor 4 m higher in 5 seconds. Whats her power output?

Zina [86]

Answer:

Explanation:

P = mgh/t = 20(9.8)(4) / 5 = 156.8 W(atts)

6 0

2 years ago

How are electrons distributed in the molecular orbitals, when the 1s orbitals of two hydrogen atoms combine to form a hydrogen m

Evgesh-ka [11]

Answer:

2 Electrons

Explanation:

Both the Hydrogen atom will share electron equally by covalent bonding. so there will be 2 electron in the bonding molecular orbital.

3 0

3 years ago