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Kazeer [188]
2 years ago
7

Ali hypothesized that increasing fertilizer would increase plant growth. Four groups of thirty similar plants were given 0 to 15

ml of fertilizer, as shown in the graph. The change in plant height for each group over a two-week period was averaged and recorded.
Physics
2 answers:
____ [38]2 years ago
8 0
<span> The hypothesis was not supported; the data indicated that too much fertilizer can inhibit plant growth. hope i helped :D</span>
vagabundo [1.1K]2 years ago
5 0

The hypnotizes was not supported.

It can be noticed that the plants that didn't get any fertilizer tend to grow less than the plants that did get moderate amount of fertilizer, but the plants that were getting much bigger amount of fertilizer didn't performed as expected and they were not growing bigger and better than the other ones. This is because the plants have certain requirements, so if they are surpassed the plants do not get any benefit plus, but instead, it can even cause them damage.

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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the
Setler [38]
C) electrical energy is transformed into heat energy
4 0
2 years ago
Read 2 more answers
A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?
mart [117]

We use the equation of motion,

S= ut+\frac{1}{2}at^{2}

Here, S is the height, u is initial velocity and a is acceleration.

Given, S = 20 \ ft S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft}  = 6.096 \ m

As  acorn falls from tree, therefore we take the value of a = 9.8 \ m/s^2 and initial velocity u = 0.

Substituting these values in equation of motion,

6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s

Thus, the time taken by the acorn to fall 20  feet ( 6.096 m ) is 1.12 s.

5 0
2 years ago
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
A point charge Q = -400 nC and two unknown point charges, q1 and q2, are placed as shown. Point charge q1 is located 1.3 meters
nalin [4]
120 nC is the answer




Sorry if I’m wrong
6 0
2 years ago
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