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kolezko [41]
2 years ago
13

What is the independent variable

Physics
2 answers:
Arisa [49]2 years ago
8 0
Independent variable is the temperature

Dependent variable is the digestive cell activity
Ludmilka [50]2 years ago
6 0

Answer:

independent variables are variables in mathematical modeling, statistical modeling and experimental science

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In a thunderstorm, the thunder occurs at the same time as lightning. Explain how does a person hears the thunder 5s after he see
sergey [27]

Answer:

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

Explanation:

4 0
2 years ago
The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val
yaroslaw [1]

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

4 0
2 years ago
a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea
frutty [35]

Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d²  = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is  3.057m

3 0
3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

6 0
1 year ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
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