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2 years ago

Honeybees can see light in the ________ range of the electromagnetic spectrum. question 8 options:

Physics

2 answers:

ELEN [110]2 years ago

8 0

<span>ultraviolet

Have a great day!</span>

vampirchik [111]2 years ago

8 0

Your question is missing the options. I've found the complete question online. It is as follows:

Honeybees can see light in the ________ range of the electromagnetic spectrum.

A. beta

B. infrared

C. ultraviolet

D. gamma

Answer:

The correct option is C. ultraviolet.

Explanation:

Even though visible to honeybees, the ultraviolet light is not visible to human eyes because its frequency is higher, and its wavelength is shorter than our brains are capable to perceive. Ultraviolet wavelength ranges from 10nm to 400 nm, which is longer than X-rays, but shorter than visible light for us.

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Having greater biodiversity in an ecosystem is beneficial to humans because

11Alexandr11 [23.1K]

Answer:

The answer is A

Explanation:

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2 years ago

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kykrilka [37]

Answer:

Clouds form when below the dew point

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2 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on

disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$

3 0

2 years ago

If an object is projected upward from ground level with an initial velocity of 160 ft per sec, then its height in feet after t

sesenic [268]

Answer:

Time will be 5 sec

And maximum height will be 1200 feet

Explanation:

We have given that initial velocity of the object u = 160 ft/sec'

Function of height $s(t)=-16t^2+160t$

Take derivative of the distance

We know that first derivative of distance is velocity

So $v(t)=-32t+160$

At maximum height we know that velocity is zero

So $-32t+160=0$

t = 5 sec

So it will take 5 sec to reach maximum height

Now maximum height will occur at t = 5 sec

So maximum height = $16\times 5^2+160\times 5=1200feet$

7 0

2 years ago

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach

Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W' the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0

2 years ago