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Dafna1 [17]
2 years ago
13

Which of the following is not a reason why plants need water? *

Physics
1 answer:
solong [7]2 years ago
5 0

Answer:

C

Explanation:

the plant cools itself down by allowing water to evaporate from their leaves so it doesn't need water to cool down

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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
forsale [732]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
where m is the mass of the substance, Cs is its specific heat capacity and \Delta T is the increase of temperature.

If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
C_s =  \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C
6 0
3 years ago
Read 2 more answers
A block of mass 5 kg is descending with a constant velocity on an inclined plane at 30°. What is the value of the frictional for
sattari [20]

Answer:28.9N

Explanation:

F=UR R=mg R=5×10 R=50N

U=tan© U=Tan30 U=1/√(3)

F=UR F=1/√(3) ×50

F=28.9N

4 0
3 years ago
Eat the majority of your calories in the evening to fuel your sleep hou<br> True<br> False
Nataliya [291]

Answer: false

Explanation:

because

7 0
2 years ago
Read 2 more answers
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
Blizzard [7]
240 meters that's it
7 0
3 years ago
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