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2 years ago

In which of the following color gradient types does the linear gradient gets mirrored on either side of the starting point?

Physics

1 answer:

Travka [436]2 years ago

4 0

Answer: Reflected

Explanation: Color Gradient refers to the variety of position dependent (as the vary with position) colors which are used to fill a region. Gradient is a mixture of two or more colors. This blend of colors adds depth to the design.

Types of color gradient are

Linear: The color blends in a straight from starting to ending point
Radial: The color transitions from starting point to ending a circular pattern.
Angle: It glides counterclockwise around starting point.
Diamond: It makes a diamond shape from starting point and its ending point is one corner of diamond.
Reflected: In this color gradient type symmetric linear gradient gets mirrored on either side of the starting point. This type of color gradient can be used to depict a lead pipe.

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A man of mass 80kg stands next to a stationary ball of mass 4kg on a frictionless surface. He kicks the ball forward along the s

scZoUnD [109]

Momentum is conserved.

p = m₁v₁ + m₂v₂ = constant

Momentum before the collision(kick):

p = 80 * 0 + 4 * 0 = 0

Momentum after the collision:
p = 80 * v₁ + 4 * 15 = 0

Solve for v₁.

3 0

2 years ago

Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,

VARVARA [1.3K]

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.

6 0

2 years ago

A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw

navik [9.2K]

Answer: 75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7 x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 = 200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0

2 years ago

Read 2 more answers

Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li

lutik1710 [3]

Answer:

Part a)

$E = 3.66 eV$

Part b)

$\lambda = 508.5 nm$

Explanation:

Part a)

change in the energy due to decay of photon is given as

$E = h\nu$

here we know that

$\nu = 8.88 \times 10^{14} Hz$

now we have

$E = (6.6 \times 10^{-34})(8.88 \times 10^{14})$

$E = 5.86 \times 10^{-19} J$

$E = 3.66 eV$

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

$\Delta E = \frac{2}{3}(3.66 eV)$

$\Delta E = 2.44 eV$

now to find the wavelength we have

$\Delta E = \frac{hc}{\lambda}$

$2.44 = \frac{1242}{\lambda}$

$\lambda = 508.5 nm$

3 0

2 years ago

Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p

Brums [2.3K]

Answer:

$P = 3.55 \times 10^5 Pa$

Explanation:

As we know that water from the fountain will raise to maximum height

$H = 26.0 m$

now by energy conservation we can say that initial speed of the water just after it moves out will be

$\frac{1}{2}mv^2 = mgH$

$v = \sqrt{2gH}$

$v = \sqrt{2(9.81)(26)}$

$v = 22.6 m/s$

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

$P = P_0 + \frac{1}{2}\rho v^2$

$P = 10^5 + \frac{1}{2}(1000)(22.6^2)$

$P = 3.55 \times 10^5 Pa$

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2 years ago