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3 years ago

Which part of a scientific argument must be supported by valid data? claim evidence reasoning theory

2 answers:

7 0

This question is a bit ambiguous because all parts of a scientific argument must be supported by valid data. However, among the choices, the closest synonym to "valid data" would be evidence. Evidence is the body of facts or information that support a given idea.

kotegsom [21]3 years ago

6 0

Answer: Theory

Explanation: A claim is an interpretation based on the results of a given experiment, you first observe, and interpret the information.

Evidence is the valid data itself.

The reasoning is the approach or the "logic path" you used to reach some theory or claim.

The theory is an interpretation of the natural world, that explains why and or how some things work.

If we want to support a theory, we need to have some valid data that supports, at least, part of it, so the right answer here is theory.

Notice that claim also needs to be supported for valid data, but in the case of the claim you first get the data and then make the claim, and in the theory, the work is done in the other order.

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A Man Moved first a Distance of 1000 m in 25 second and 2.5 km in 50 second along a in straight line?

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Answer:

Average speed = 46.67 m/s

Explanation:

Given that the time taken in covering first 1000 m = 25 seconds.

The time taken in covering next 2.5 km = 50 seconds.

Total distance covered = 1000 m + 2500 m = 3500 m

Total time taken = 25+50=75 seconds

Average speed = Total distance covered / total time taken

= 3500/75 = 46.67 m/s

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Read 2 more answers

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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3 years ago

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

Tema [17]

In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

3 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

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3 years ago