2 years ago

6. Given that white light contains all colors of the spectrum, what growth result would you expect under white light?

1 answer:

4 0

You could probably expect normal plant growth, as a white light is similar to the sun in the respect that it contains all colors of the spectrum.

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Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of

igomit [66]

Answer:

T=Lnsin $\alpha$

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin $\alpha$ (nsin90)

T=Lsin $\alpha$ xn

T=Lnsin $\alpha$

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2 years ago

a feather is dropped on the moon from a height of 1.40meters. the acceleration of gravity on the moon is 1.67m/s^2. determine th

kykrilka [37]

Answer:

1min since there is no gravity on the moon so it will take time to drop on the moon.

Explanation:

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2 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

2 years ago

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Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

2 years ago

By using fossils and matching layers rock layers can be correlated to eachother.

bija089 [108]

I think that the answer to that is true hope that helps

4 0

3 years ago

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