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3 years ago

Please helppp nowwwwww

Physics

1 answer:

kogti [31]3 years ago

5 0

Answer:

4.) the dependent variable is the french fry

5.) the independent variable is the jar; whether it has a lid on or not

6.) there are plenty of control variables in this example, the type of jar (size, material of the jar, etc.), the type of french fry, the room that the jars are in, the air pressure in the room, how long the experiment is, etc.

7.) My conclusion would be that the french fry that has the most exposure to the air.

The moisture in the french fry is leaving the jar because it is exposed to the open air. It becomes stale. But because the other french fry is in the closed jar, it is not exposed to the air or any new type of bacteria that could've entered the open jar. It'll stay fresher longer because it is not exposed the way the fry in the open jar is.

Explanation:

The dependent variable is the variable that is being tested. It is the one that changes due to whatever is being placed upon it.

i.e. the dependent variable in an experiment that tests the growth rate of lima bean plants in different colored lighting is the lima bean plant. <em>Depending</em> on the color of the lighting, the lima bean plant will grow at different rates.

The independent variable is the variable that is changed or controlled by an experimenter

i.e. the independent variable in an experiment that tests the growth rate of lima bean plants in different colored lighting is the different colored lighting. The color of the lighting helps prove the hypothesis that the growth rate of a lima bean plant is dependent on the lighting. In other words, the color of the light is not dependent on the growth of the plant.

A control variable is a variable that cannot be changed and is constant throughout the entire experiment

i.e. temperature, gravity, pollutants in the air, air pressure, humidity, etc.

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3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

4 years ago

Using a mass of 1. 8 g and the volume displaced by the sample, calculate the sample's density. Question 17 options: 0. 17 g/mL 0

LUCKY_DIMON [66]

The sample's density with respect to the volume it displaces is equal to: C. 0. 60 g/mL

<u>Given the following data:</u>

Mass = 1.8 grams

Initial volume = 20 mL

Final volume = 23 mL

To calculate the sample's density:

First of all, we would determine the volume displaced by the sample;

$Volume \;displaced = 23-20$

Volume displaced = 3 mL

Density can be defined as mass all over the volume of an object. Thus, density is mass per unit volume of an object.

Mathematically, the density of a substance is given by the formula;

$Density = \frac{Mass}{Volume}$

Substituting the given parameters into the formula, we have;

$Density = \frac{1.8}{3}$

Density = 0.60 g/mL

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3 years ago

How are the angles of incidence and reflection related at a reflection surface?

Citrus2011 [14]

Answer:

The angle of incidence is equal to the angle of reflection.

Explanation:

yes.

3 0

3 years ago