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3 years ago

Hybrid plants usually have increased Vigar true or false

Physics

1 answer:

tester [92]3 years ago

3 0

Answer: True

Explanation:

Because plants have an increase in vigar to grow

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Professor blossom expects her students to be on time for class holds, them for the entire class period and has firm deadlines se

defon

Answer:

A dictatorship or tyranny.

Explanation:

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3 years ago

How big is the planet earth

AnnZ [28]

After doing some research on Wiki, Google, and Easyscienceforkids, I have come to a conclusion that:

Mass: 5.972 * 10^24 kg

Radius: 3958.8 mi

Distance from the sun: 92.96 million mi

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3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

5 0

3 years ago

What is the value of a in the function’s equation?

Dennis_Churaev [7]

Answer:

Explanation:

What is the value of a in the function's equation?

A. -2

B. -3

C. 3

D. 2

ANSWER: c.

6 0

2 years ago

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

TiliK225 [7]

Answer:

$E = 6.77 *10^{3} N/C$

Explanation:

THE GIVEN sheet can be taken as two horizontal force with surface charge density is

at one surface is ∈_1 = $95*10^ {-9} C/mm2$

at oher surface is ∈_2= $-25*10^{-9} C/mm2$

the magnitude of electric field due to surface charge is given as $\frac{∈}{2∈_O}$

So, electric field at P (2 CM below from surface is) = E_1 +E_2

$E = \frac{∈_1}{2∈_O} +\frac{∈_2}{2∈_O}$

$E = \frac{95*10^{-9} +25*10^{-9}}{2*8.85*10^{-12}}$

$E = 6.77 *10^{3} N/C$

8 0

4 years ago