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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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A hot–air balloon is moving at a speed of 10 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–dir
IrinaVladis [17]
The ball has an initial speed of 10m/s. This is because it is moving with the balloon. Now the balloonist throws the ball 4m/s with respect to himself, so it means that he gives the ball a extra push of 4m/s, so the total speed is 14m/s. Since it takes 30 seconds to reach the ground, the distance travelled is 14*30=420m.
7 0
2 years ago
A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0
2 years ago
What is Otter's average velocity over his entire trip when it takes him 2 minutes to walk 100 meters north and another 1 minute
Leya [2.2K]

Answer:

0.50m/s

Explanation:

Average velocity is the change in displacement of a body with respect to time.

Velocity = ∆S/∆t

∆S = 100m - 70m

∆S = 30m

∆t = 2min - 1 min

∆t = 1min = 60secs

Substitute the given parameters into the formula for velocity

Velocity = 30m/60s

Velocity = 1/2 m/s

Average Velocity = 0.5m/s

7 0
2 years ago
Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose
dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by h=7\ m

Suppose the dolphin jump with an initial velocity of u

so u is given by u^2=2\cdot g\cdot h

If dolphin launches at an angle \theta then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

R=\frac{u^2\sin 2\theta }{g}

substitute the value of u^2

R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}

R=2h\sin 2\theta

Range will be maximum for \theta =45^{\circ}

thus R_{max}=2\times 7\times 1=14\ m

                                     

3 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
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