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2 years ago

6

What is the mean height of the plants given no fertilizer?

1 answer:

Marianna [84]2 years ago

3 0

Answer:

B. 33.5 cm

Explanation:

Steps:

Add all numbers, and divide by the number of addends.

Add 30.5, 32.1, 40.6, 30.5,and 33.8, and you get 167.5

Divide them by 5, and you get 33.5

You might be interested in

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago

A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.

Alex Ar [27]

Good. You can do some very interesting experiments with that equipment.

3 0

3 years ago

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i

scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0

3 years ago

3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl

Marina86 [1]

The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.

6 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago