All categories

3 years ago

Where are the nodules of a legume found?

Physics

1 answer:

Sveta_85 [38]3 years ago

6 0

Answer:

The answer is roots.

Explanation:

Root nodules are found on the roots of plants, primarily legumes, that form a symbiosis with nitrogen-fixing bacteria. so it is roots

You might be interested in

Help me please!!!

lilavasa [31]

Both planets are similar in shape and have a rocky surface. Not sure about the phases though

7 0

3 years ago

Read 2 more answers

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

5 0

3 years ago

If you were to double the separation between two slits, by what factor would the number of interference fringes within the centr

NemiM [27]

Answer The fringes become closer together as the slits are moved farther apart.

4 0

3 years ago

Which statement about convection is correct?

AlekseyPX

D. Convection occurs when heated particles of a material flow toward areas having less thermal energy. This movement of particles can only occur in gases and liquids, not solids.

7 0

3 years ago

Read 2 more answers

A woman and her dog are out for a morning run to the river, which is located 4.0 KM away. The woman runs at 2.5 M/S in a straigh

VLD [36.1K]

River shore is located at distance

$d = 4 km$

speed of the woman is given as

$v_1 = 2.5 m/s$

now the time taken by the woman to cover the distance is

$t = \frac{d}{v}$

$t = \frac{4000}{2.5} = 1600 s$

for the same time interval the dog will run to and fro with speed 4.5 m/s

so the total distance moved by the dog is given by

$d = v* t$

$d = 4.5 * 1600$

$d = 7200 m$

so the total distance that dog will move is 7200 m or 7.2 km

8 0

3 years ago