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3 years ago

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Sanjeev noticed his plants were wilting and turned on the sprinklers in his garden. An hour later, he notices that the plant's l

eaves aren’t wilted any longer. Why did this happen?
A) The xylem stored the water in the stem of the plant, making it more rigid

B) The water flowed out of the vacuole creating turgor pressure in the plant’s cells

C) The plant’s stored energy was activated by the water

D) The water entered the vacuole creating turgor pressure in the plant’s cells

help please ASAP

2 answers:

sineoko [7]3 years ago

8 0

Answer:

D

Explanation:

alexira [117]3 years ago

3 0

Answer:

B

Explanation:

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What does physical education mean

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4 years ago

Read 2 more answers

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.

Darina [25.2K]

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $\frac{2050}{150*10^{-3}*15} = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

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3 years ago

Help me please I’m so bad at physics

aniked [119]

Answer:

C would probably be the best answer.

Explanation:

Switches control the flow of an electrical current in a circuit. Switches open or close the circuit. If you open a circuit, you turn it off, and if you close it, you turn it on.

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2 years ago

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

melomori [17]

Answer:

This is because below 4°c, water unlike other materials becomes less dense when it's temperature is further lowered.

Explanation:

Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

This behavior is due to the hydrogen bonding of the water molecules.

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3 years ago

A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that,

amm1812

Answer:

$\Delta V=0.0592\ L$

Explanation:

Given:

Initial temperature of the coolant, $T_f=95^{\circ}C$

final temperature of the coolant, $T_f=105^{\circ}C$

total volume of the coolant, $V_c=16\ L$

coefficient of volume expansion for coolant, $\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}$

volume of Al radiator, $V_a=2\ L$

volume of steel radiator, $V_s=14\ L$

<u><em>We have:</em></u>

coefficient of volume expansion for Aluminium, $\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}$

coefficient of volume expansion for steel, $\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}$

<u>Now, change in volume of the coolant after temperature rises:</u>

$\Delta V_c=V_c.\beta_c.\Delta T$

$\Delta V_c=16\times 410\times 10^{-6}\times (105-95)$

$\Delta V_c=0.0656\ L$

<u>Now, volumetric expansion in Aluminium radiant:</u>

$\Delta V_a=V_a.\beta_a.\Delta T$

$\Delta V_a=2\times 75\times 10^{-6}\times (105-95)$

$\Delta V_a=0.0015\ L$

<u>Now, volumetric expansion in steel radiant:</u>

$\Delta V_s=V_s.\beta_s.\Delta T$

$\Delta V_s=14\times 35\times 10^{-6}\times (105-95)$

$\Delta V_s=0.0049\ L$

∴Total extra accommodation volume created after the expansion:

$V_X=\Delta V_s+\Delta V_a$

$V_X=0.0049+0.0015$

$V_X=0.0064\ L$

Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.

$\Delta V=\Delta V_c-\Delta V_X$

$\Delta V=0.0656-0.0064$

$\Delta V=0.0592\ L$

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3 years ago