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3 years ago

Write the name of any two plant hormones used by use for tissue culture

Physics

1 answer:

Mice21 [21]3 years ago

3 0

Answer:

The plant hormones auxin and cytokinin are critical for plant regeneration in tissue culture, with cytokinin playing an instrumental role in shoot organogenesis.

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Can someone help me with these two please.

Alex17521 [72]

Could you scroll up a little so I can see the first part of the question- I MIGHT be able to help you then. Thanks!

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3 years ago

Read 2 more answers

What are two properties that lead to ElectroMagnetic Interactions? please help this is due very soon!!!

melisa1 [442]

Answer:

Electric charge. A fundamental property that leads to the electromagnetic interactions among particles that make up matter.

7 0

3 years ago

Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0

3 years ago

In the figure , if Q = 30 uC , q = 5.0 uC , and d = 30 cm , what is the magnitude of the electrostatic force on g?

lana [24]

Answer:

$F = k\frac{q_1q_1}{ {d}^{2} } \\ = 9.0 \times {10}^{9} (N {m}^{2} {C}^{ - 2}) \frac{30 \times {10}^{ - 6} (C)5.0 \times {10}^{ - 6} (C)}{ {0.3}^{2} } N$

3 0

2 years ago

A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength

Step2247 [10]

Answer:

Magnetic field, $B=4.16\times 10^{-5}\ T$

Explanation:

It is given that,

Velocity of proton, $v=3.5\times 10^7\ m/s$

Magnetic force, $F=1.65\times 10^{-16}\ N$

Charge of proton, $q=1.6\times 10^{-19}\ C$

We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :

$F=qvB\ sin\theta$

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}$

B = 0.0000416 T

$B=4.16\times 10^{-5}\ T$

Hence, this is the required solution.

4 0

3 years ago

Write the name of any two plant hormones used by use for tissue culture​

Write the name of any two plant hormones used by use for tissue culture