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3 years ago

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James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air-conditione

d rooms. He then measured the growth of the pea plants after a week. The growth of pea plants in this experiment is a(n) __________. Question 7 options: A) constant B) dependent variable C) control D) independent variable

1 answer:

lord [1]3 years ago

6 0

Dependent variable: it is dependent on the amount of light received.

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A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

5 0

2 years ago

In the mixtures lab, you waited to see if the liquid would form lumpy or fluffy masses, which would indicate it was a _____.

Over [174]

B) colloid because i took the test and that the answer

5 0

3 years ago

Read 2 more answers

Vladimir79 [104]

Answer:

pagtalon?

Explanation:

because that had to be the answer

5 0

2 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

2 years ago

An object is held at rest and allowed to drop How far does the object fall in 3.3 seconds?

IgorC [24]

After x seconds, an object will fall $\frac{1}{2}at^2$ where a is acceleration due to gravity and t is time

so when t=3.3
the distance it will fall is $(\frac{1}{2})(9.8\frac{m}{s^2})(3.3m)^2$ =53.361m
it will fal 53.361 meters

6 0

2 years ago