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3 years ago

Which part of a flowering plant attracts insects, birds, and small mammals?

Physics

2 answers:

mezya [45]3 years ago

4 0

Answer: petals or the scent of a flower

Explanation:

because they are brightly colored it attracts pollinators. many flower produce a sweet liquid called nectar which is what insects feed on. so that also attracts them

madam [21]3 years ago

3 0

The flowers attracts insects, birds, and small animals. They also attract humans, who admire the

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Which equations represent the relationship between wavelength and frequency for a sound wave? Check all that apply. V = f = vf f

Vikentia [17]

It's really hard to make out the gibberish at the end, where the list of choices should be.

Here are the three equations for the relationships among the wavelength, frequency, and wave-speed, for ANY wave:

-- Speed = (wavelength) x (frequency)

-- Wavelength = (Speed) / (frequency)

-- Frequency = (Speed) / (wavelength)

Notice that you only have to remember any ONE of these. Then you can always get the other ones by multiplying or dividing both sides by the same quantity.

If you want to remember all of them, here are the short forms:

V = L · f

L = V / f

f = V / L

6 0

3 years ago

Read 2 more answers

Imagine we cook an egg by immersing it into water which is boiled by an electric heater. The heater utilizes a current I =10 A a

tatuchka [14]

Answer:

Q' = 115.2 KJ

Explanation:

Given that

Current I = 10 A

Voltage = 120 V

Time t= 120 s

The energy supplied Q= V I t

Q= 10 x 120 x 120 J

Q= 144 x 1000 J

Q= 144 KJ

The change in the energy ΔEngg = 28.8 kJ

By using energy conservation

Q= Q' + ΔEngg

Q'=Wasted energy

Now by putting the values in the above equation

144 = Q' + 28.8 KJ

Q' = 115.2 KJ

Therefore the waste energy will be 115 kJ.

8 0

3 years ago

a merry go round has a radius of 8 meters makes 2 revolutions every 2.5 minutes. A. express the angular speed of the merry go ar

pantera1 [17]

Answer:

a). $v=83.77x10^{-3} rad/s$

b). $v=0.8$ rpm

c).v=0.5865 m/sec.

Explanation:

Given:

$r=8m$

$v=\frac{2rev}{2.5minutes}$

a).

$2 rev*\frac{2\pirad}{1rev}=4\pi rad$

$t=2.5minutes*\frac{60s}{1minute}=150s$

The angular speed in radians per seconds is

$v=\frac{4\pi}{150s}=83.77x10^{-3} rad/s$

b).

$v=\frac{2rev}{2.5minute}rpm$

$v=0.8$ rpm

Child's distance per revolution

(pi*2r) = 43.988 metres.

v=(43.988 x 0.0133333) = 0.5865 m/sec.

4 0

3 years ago

A syrup added to an empty container with a mass of 115.25g . When 0.100 pint of syrup is added, the total mass of the container

sesenic [268]

Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. We calculate as follows:

Density = 182.48 g - 115.25 g / 0.100 pint ( 0.47 L / 1 pint ) = 67.23 g/0.047L
Density =1430.43 g/L

3 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago