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3 years ago

If an object reflects all of the wavelengths of visible light you would see what color

Physics

1 answer:

Ahat [919]3 years ago

8 0

Answer:

white

An object has the color of the wavelengths it reflects. A material that reflects all wavelengths of visible light appears white. A material that absorbs all wavelengths of visible light appears black. A green lime absorbs most wavelengths but reflects green, so the lime looks green, as shown below.

Explanation:

hope that helps.

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What Are Three Characteristic Properties Of Ionic Compounds

sashaice [31]

Ionic bonds have...

- High melting points
- Conductive properties
- Are good insulators
(Extra: Form crystal-like structures over just plain molecules!)

7 0

3 years ago

Remember to include your data, equation, and work when solving this problem.

andrezito [222]

Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\$

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

$F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]$

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

6 0

3 years ago

Amina has a mass of 48.4kg, what is her weight.please show calculation

Masja [62]

G=m×g

since g=10N,

G=48.4×10

G=484

I hope it's true .d

good luck

6 0

3 years ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$