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3 years ago

HELP! DUE TOMORROW!! specify the independent and dependent variables, as well as constants. 4-6! only!

Physics

1 answer:

olga55 [171]3 years ago

5 0

Answer:

the variables in a study of a cause-and-effect relationship are called the independent and dependent variables. The independent variables is the cause and the dependent variables is the effects. It's value depends on the changes in the independent variable.

Explanation:

I hope this small infor is helpful

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A supply plane needs to drop a package of food to scientists working on a glacier in greenland. the plane flies 100 m above the

pishuonlain [190]

The package should be dropped 678 m short of the target.

A package dropped from a plane which is moving at a speed v, has a horizontal velocity equal to the horizontal velocity of the plane. It has a parabolic trajectory, traversing a horizontal range x while it falls through a vertical height y.

The package has no initial vertical velocity, and it falls through a height y under the action of the acceleration due to gravity g.

Use the equation,

$y=\frac{1}{2} gt^2$

Write an expression for t, the time taken for the package to fall through y.

$t^2=\frac{2y}{g}$

Substitute 100 m for y and 9.81m/s² for g.

$t^2=\frac{2y}{g}\\ =\frac{2(100m)}{9.81m/s^2} \\ =20.39s^2\\ t=4.52s$

In the time t the package travels a horizontal distance x. The horizontal velocity of the package remains constant, since no force acts along the horizontal direction.

Therefore, the horizontal distance traveled by the package is given by,

$x=vt\\ =(150m/s)(4.52s)\\ =678m$

If the package is released 678m before the target, the package would reach the scientists working in Greenland.

6 0

4 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0

4 years ago

silicon is the principal component of glass and has properties of both metals and non-metals. this makes silicon A) Metal B) Non

Eddi Din [679]

Silicon is in fact a Metalloid. Comment if you need more info!

5 0

3 years ago

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

8 0

3 years ago