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3 years ago

What is a secondary color of light?

Physics

2 answers:

jasenka [17]3 years ago

5 0

It is Yellow, Cyan, Magenta . The answer is D.

k0ka [10]3 years ago

3 0

C. It is a combination of two primary colors.

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An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t

Dimas [21]

Answer:

6.5 seconds.

Explanation:

Given: h=-16t²+679

When the object reaches the ground, h=0.

∴ 0=-16t²+679

collecting like terms,

⇒ 16t²=679

Dividing both side of the equation by the coefficient of t² i.e 16

⇒ 16t²/16 = 679/16

⇒ t² = 42.25

taking the square root of both side of the equation.

⇒ √t² =√42.25

⇒ t = 6.5 seconds.

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4 years ago

When an object is moving with uniform circular motion, the centripetal acceleration of the object a. is circular. b. is perpendi

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When an object is moving with uniform circular motion, the centripetal acceleration of the object d. is directed toward the center of motion.

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3 years ago

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Please help me easy question just need help

Nesterboy [21]

Answer:

what is Jose potential energy

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3 years ago

What occurs to the arrangement of water molecules as water melts

Nikolay [14]

The arrangements of the water molecules they begin to spread apart. As the water(ice) melts the molecules develop fewer compact and start juddering harder until the ice has changed state into liquid.

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3 years ago

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A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00

marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

$W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)$ (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

$-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}$

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

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4 years ago