Which one of the following is not a safety rule in the laboratory. a. To sit on a stool
b. To follow instructions
c. To wear shoes with hard soles
d. Handling apparatus

Answers

Answer 1

Answer:

Answer:

C. To wear shoes with hard soles

Explanation:

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Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

Answers

Answer:

1) When $d_(o)$ < $d_(i)$ (hence $d_(o)$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_(o)$ > $d_(i)$ (and $d_(o)$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$(1)/(f)=(1)/(d_(o)) + (1)/(d_(i))$

$Magnification, \, m = (h_(i))/(h_(o)) = -(d_(i))/(d_(o))$

Where:

f = Focal length of the mirror = R/2

$d_(i)$ = Image distance from the mirror

$d_(o)$ = Object distance from the mirror

$h_(i)$ = Image height

$h_(o)$ = Object height

$d_(o)$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_(i)$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_(o)$ < $d_(i)$ (hence $d_(o)$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_(o)$ > $d_(i)$ (and $d_(o)$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

PLEASE HELP ON MORE SCIENCE! :)

Answers

1 is standing wave
2 is Constructive Interference
3 is longitudinal
4 is first choice
5 is first choice

The second one: "Constructive interference" even though the question confused me

Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane.

Answers

The moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Given data:

The mass of each sphere is, $m = 0.200 \;\rm kg$ .

Length of side of square is, $L = 0.400 \;\rm m$ .

The expression for the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is,

$I = 4 mR^(2)$

Here,

R is the distance between center of the square and the sphere. And its value is,

$R =(1)/(2)\sqrt{L^(2)+L^(2)}\nR =(1)/(2)\sqrt{0.400^(2)+0.400^(2)}\nR = 0.282 \;\rm m$

Then, moment of inertia is,

$I = 4 mR^(2)\nI = 4 * 0.200 * 0.282^(2)\nI = 0.0636 \;\rm kg-m^(2)$

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Learn more about moment of inertia here:

brainly.com/question/2176093?referrer=searchResults

The moment of inertia of the system about an axis through the center of the square, perpendicular to the plane is 0.064 kg.m²

$\texttt{ }$

Further explanation

Let's recall Moment of Inertia formula as follows:

$\boxed{ I = m R^2 }$

where:

I = moment of inertia

m = mass of object

R = distance between the object and the axis of rotation.

Given:

mass of sphere = m = 0.200 kg

length of side = x = 0.400 m

Asked:

net moment of inertia = ΣI = ?

Solution:

Let's ilustrate this question as shown in the attachment.

Firstly , let's find distance between center of the square and the sphere:

$R = (1)/(2) √(x^2+x^2)$

$R = (1)/(2) √(2x^2)$

$R = (1)/(2)√(2) x$

$R = (1)/(2) √(2) (0.400)$

$\boxed{R = 0.200√(2) \texttt{ m}}$

$\texttt{ }$

Next , we could find total moment of inertia as follows:

$\Sigma I = mR^2 + mR^2 + mR^2 + mR^2$

$\Sigma I = 4mR^2$

$\Sigma I = 4(0.200)(0.200√(2))^2$

$\boxed{\Sigma I = 0.064 \texttt{ kgm}^2}$

$\texttt{ }$

Learn more

Ratio of Moment of Inertia : brainly.com/question/2176655
Net Torque on the objects : brainly.com/question/9612563
Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

According to uniformitarianism, which statement is true? A. Weathering, erosion, and uplift occur slower now than in the past. B. Weathering, erosion, and uplift occur faster now than in the past. C. Weathering, erosion, and uplift occur at the same rates. D. Weathering, erosion, and uplift occur at the same rates now as they did in the past.

Answers

c because uniformitarianism follows the belief that changes have always happened at a constant rate

According to the first law of thermodynamics what could happen when heat is added to a system

Answers

Answer: Increase internal energy and work done by the system

Explanation: According to the first law of thermodynamics conservation of energy, when heat is added to the system then the system gain or loss of energy.

So, when we add heat to a system then increases internal energy and external work done by the system.

temperature rises ? specific heat capacity etc ...

If a turtle walks at a constant speed of 0.23 m/s, how long will it take to cover a distance of 65 meters?

Answers

Time = (distance) / (speed)

Time = (65 meters) / (0.23 m /s)

Time = (65 / 0.23) sec

Time = 282.6 seconds ( that's 4 minutes 42.6 seconds)

Final answer:

The time it would take a turtle moving with a speed of 0.23 m/s to cover a distance of 65 meters is approximately 282.61 seconds.

Explanation:

The subject of this question is physics, specifically the concept of speed, distance, and time. The formula that relates these three quantities is speed = distance/time.

In this case, we are asked to calculate the time it would take for a turtle moving with a constant speed to cover a certain distance. Here, the given speed (0.23 m/s) of the turtle and the distance it needs to cover (65 meters) are known.

By rearranging the formula to solve for time, we get time = distance / speed. Plugging in the values, we get time = 65 meters / 0.23 m/s. Doing the math gives approximately 282.61 seconds. Thus, it would take the turtle around 283 seconds to cover the distance of 65 meters at a speed of 0.23 m/s.