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ololo11 [35]
4 years ago
8

Becky wanted to figure out what type of liquid worked best for growing beans . She wanted one with coca-cola ,one with lemonade,

and one with just water. After one week,she measured how high that had grown.
Physics
1 answer:
pashok25 [27]4 years ago
4 0
Water is the best liquid hope this helps 
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Can I get the answer
ANEK [815]
Hey there
how are u..................................................
4 0
3 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
Two forces that are not equal in size are
Bumek [7]
They are unbalanced forces ..... Hope this helps :3
4 0
3 years ago
Perpetual Endless Relentless Stubborn Immovable Steady Tireless Enduring Never-ending Constant Ever-lasting
klasskru [66]

Eleven truly inspiring adjectives !
You've made my day.

Do you have a question to ask ?
5 0
3 years ago
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