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2 years ago

What relationship is flowers and honey bees

1 answer:

7 0

A symbiotic relationship. One where two species benefit each other. In this case, the flower gives the bee nectar, so the bee pollinates the flower.

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¿Cuál es la frecuencia de una ola con una velocidad de 14 m / s y una longitud de onda de 20 metros?

Margaret [11]

Responder:

<h2>0.7Hertz </h2>

Explicación:

Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.

Velocidad de una onda = frecuencia * longitud de onda

v = fλ

Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros

frecuencia f = v / λ

f = 14/20

f = 0.7Hertz

La frecuencia de la onda es de 0.7 Hertz.

7 0

2 years ago

List five situations in which a lesser amount of friction would be beneficial. Overall, would a high or low coefficient of frict

Natalka [10]

Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

6 0

2 years ago

Read 2 more answers

What velocity does a 2kg mass have when its kinetic energy is 16 J

alexandr402 [8]

We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v.

K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s

Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!

7 0

2 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

2 years ago

Explain the terms boiling and boiling point.

Volgvan

The particular temperature at which vaporisation occurs is known as the boiling point of liquid. Volume of water increases when it boils at 100° C. 1 cm3 of water at 100 ° C becomes 1760 cm3 of steam at 100 ° C.

Hope it helps!!!!!!!!!!!!!!!!!!!!! ~~~~~~~~~~~~~~~~~~~

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3 0

2 years ago