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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 ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0
2 years ago
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The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light
Nikolay [14]
<span>A. Man Ray --------------------</span>
7 0
1 year ago
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Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

7 0
2 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
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