When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

Answers

Answer 1

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

Related Questions

The main purpose of the turbine in the turbojet engine is to A. drive the compressor.B. reduce the temperature of the exhaust gas.C. increase the velocity of the exhaust gases.D. compress the air. ~IT is NOT D.
Dua buah vektor, yaitu P= 6i-5j+2k dan Q = 2i+2j+8k. Tentukan vektor R agar P-Q+R=0. Hitung pula besar vektor R?
In your own words, describe the matter found in the universe and the relationship between stars, galaxies, and planets.
Support that P(e)=0.76,P(F)=0.45,and P(E and F)=0.32. What is P (E or F)?
Is it possible for a population with a high birth rate to decrease in size.

How do you connect 1 ohm, 2 ohm and 3 ohm resistors to getan equivalent resistance of 5/6 ohm.

Answers

You (or your homework sheet) may be fishing for the answer "connect
the 3 resistors in parallel", but that does not do the job.

When those 3 resistors are connected in parallel, their combined resistance
is 6/11 ohm.

If the 2 and the 3 alone are in parallel, their combination is 6/5 ohms, and
adding the 1 ohm in series makes 2.2 ohms.

If the 1 and the 2 alone are in parallel, their combination is 2/3 , and adding
the 3 in series makes 3 and 2/3 .

If the 1 and the 3 alone are in parallel, their combination is 3/4 , and adding
the 2 in series makes 2.75 .

There is nothing you can do with those 3 resistors, or with any 2 of them,
or with any 1 of them, that will give you 5/6 ohm .

If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, what's the average velocity of the bodyA-17 m/s
B-19.5 m/s
C-28 m/s
D-21.5 m/s

Answers

If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, the average velocity of the body is 21.5 m/s. the answer is letter D. Velocity is a vector quantity and average velocity can be defined as the sum of the initial and final velocity divided by 2 at constant acceleration.

The Correct answer to this question for Penn Foster Students is: 21.5 m/s

Is an object moving in uniform circular motion accelerating?

Answers

Yes.

-- 'Acceleration' does NOT mean 'speeding up'.
It means ANY change in the speed OR direction of motion ...
speeding up, slowing down, or turning.

-- If an object is NOT moving in straight line at constant speed,
then its motion is accelerated.

-- In circular motion, or even just going around a curve,
the object is accelerating, because its direction is constantly
changing, even if its speed is constant.

Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200W/(m⋅K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.60m in diameter having a layer of fat 3.90cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.80∘C during hibernation, while the inner surface of the fat layer is at 30.9∘C a) What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W ? b) How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.4W ?

Answers

A) The temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W is; T_i = 38.52°C

B) The thickness of the layer contained within the fur so that the bear loses heat at a rate of 51.4 W is; t = 13.41 cm

We are given;

Diameter of sphere; d = 1.6 m

Radius of sphere; r = d/2

r = 1.6/2

r = 0.8 m

Thickness of bear; t = 3.9 cm cm = 0.039 m

Outer surface Temperature of fur; T_h = 2.8 ∘C

Inner surface Temperature of fat;T_f = 30.9 ∘C

Thermal conductivity of fat; K_f = 0.2 W/m⋅k

Thermal conductivity of air; K_a = 0.024 W/m⋅k

A) To find the temperature at the fat-inner fur boundary when heat loss is 51.4 W, we will use the heat current formula;

H = K_f•A(T_f - T_i)/t

Where;

A is area = 4πr²

A = 4π × 0.8²

A = 8.04 m²

T_i is the temperature we are looking for

H is heat loss = 51.4

t is thickness

Making T_i the subject gives;

T_i = (T_f × H × t)/(K_f × A)

T_i = (30.9 × 51.4 × 0.039)/(0.2 × 8.04)

T_i = 38.52°C

B) We want to find the thickness of the layer contained within the fur. Thus, we will use K_a instead of K_f. Let us make t the subject in the heat current formula to get;

t = (K_a•A(T_i - T_h)/H

t = (0.024 × 8.04 × (38.52 - 2.8))/51.4

t = 0.1341 m

t = 13.41 cm

Read more at; brainly.com/question/14548124

Answer:

Explanation:

Using the equation

H = Q/t = k A ( T hot - T cold) / L

where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m

radius = 1.60 m / 2 = 0.80 m

A = 4 × 3.142 × ( 0.8²) = 8.04352 m²

making T cold subject of the formula

T cold = T hot - $(HL)/(KA)$ = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) = 30.9° C - 1.25 ° C = 29.65° C

b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W

thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula

L = $\frac{kA( T HOT - T COLD )}H}$ = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m

If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscillations. What can we say, about these springlike bonds when the charge is first, furthest away and second, closest to the source charge

Answers

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second and closest to the source charge. The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source when Furthest away the bond with charge will be less effective

Alpha particles \ beta particles / gamma radiation Put these different types of radiation in order from MOST to LEAST penetrating.

Answers

It goes: Gamma radiation > beta particles > alpha particles

Answer:

Gamma > Beta > Alpha

Explanation:

Other Questions

How many kilograms are in 12321 grams?

How many g in 6 1/2 kg

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 80 kg, are pulled at a constant speed by a tow rope that maintains a constant tension of 360 N. How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Function and location of esophagus