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3 years ago

Can any one please help

Physics

2 answers:

marta [7]3 years ago

5 0

Answer:

1. A) lamb

candle

torch

sun

B) the stars

light bulb

Toxic Glow-in-the-Dark Millipedes

C) no, humans are not luminous objects

D) on a sunny day the light from the sun hits us and then the light gets reflected to the friends eyes and that is how they can see us.

Explanation:

Alinara [238K]3 years ago

3 0

Answers:

A)
- lamp
- candle
- flashlight
- the sun

B)
- one in space: the moon
- one in your home: light bulb
- one that is a living creature: firefly

C)
- non- luminous

D)
- A friend can see you on a sunny day because the sun is shining on them. The sun creates light so you are able to see them.

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If a ball starts with the velocity of 2.0m/s and accelerates with a constant rate of 0.50m/s squared for 2.5s what is the balls

baherus [9]

Acceleration can be any change in speed, increasing or decreasing.
You haven't said whether the ball is speeding up or slowing down.

If its acceleration is positive ... speed is increasing ... then in 2.5 seconds,
it GAINS (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed. Added to its initial
speed of 2.0 m/s, it ends up moving at 4.5 m/s.

If its acceleration is negative ... speed is decreasing ... then in 2.5 seconds,
it LOSES (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed. Added to its initial
speed of 2.0 m/s, it ends up moving at -0.5 m/s. That means that it ends up
moving in the opposite direction compared to its direction at the beginning of
the change.

7 0

3 years ago

What is reflection of light

AnnZ [28]

Answer: When a ray of light approaches a smooth polished surface and the light ray bounces back, it is called the reflection of light. The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray.

Have a great day and stay safe !

4 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$

$f_2 = n+f_1$

Replacing $3/s$ for n and 202Hz for $f_1,$

$f_2 = 3/s + 202Hz$

$f_2 = 3/s(\frac{1Hz}{1/s})+202Hz$

$f_2 = 206Hz$

The frequency of the tightened is 205Hz

7 0

3 years ago

Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.

MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0

3 years ago

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

3 years ago