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slava [35]
3 years ago
11

If the temperature of a liquid increases, the density of the liquid decreases because the particles move farther apart. What are

the variables in her hypothesis?
Physics
1 answer:
Lorico [155]3 years ago
8 0

This question is incomplete because part of the content is missing; here is the complete question:

Ingrid wrote the hypothesis below.  If the temperature of a liquid increases, the density of the liquid decreases because the particles move farther apart.

What are the variables in her hypothesis?

A. The independent variable is the temperature, and the dependent variable is the density.

B. The independent variable is the density, and the dependent variable is the temperature.

C. The independent variable is the temperature, and the dependent variable is the distance between particles.

D. The independent variable is the distance between particles, and the dependent variable is the temperature.

The answer to this question is A. The independent variable is the temperature, and the dependent variable is the density.

Explanation:

To begin, the variables in an experiment are the factors being studied or analyzed, which are expressed in the hypothesis. According to this, the two variables in the experiment are the temperature and the density.

Additionally, in experiments, it is common one of the variables is independent, which is the factor manipulated by the researcher, and the other is dependent as this is expected to be changed by the first variable. If this is applied to the experiment described, the temperature is independent because this will be manipulated by Ingrid to prove density changes. Also, the density is expected to be affected by the first variable, and therefore density is the dependent variable.

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kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
Melanie completes a long distance run at an average speed of 6 mph. If it takes her 3 hours, how far did she run?
Rama09 [41]

Answer:

18 miles

Explanation:

The average speed is 6 mph

Melanie ran for 3 hours

Speed × Time = Distance

So, 6 mph × 3 h = 18 miles

6 0
4 years ago
A racecar traveling at a velocity of 18.5 m/s, accelerates at a rate of 2.47 m/s2 and covers a distance of 79.78 m. Determine th
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Answer:

The final velocity of the race car is 27.14 m/s

Explanation:

Given;

initial velocity of the race car, u = 18.5 m/s

acceleration of the race car, a = 2.47 m/s²

distance covered by the race car, s = 79.78 m

Apply the following kinematic equation to determine the final velocity of the race car.

v² = u² + 2as

v² = (18.5)² + 2(2.47)(79.78)

v² = 736.363

v = √736.363

v = 27.14 m/s

Therefore, the final velocity of the racecar is 27.14 m/s

4 0
4 years ago
How far can light travel in one year?
elena-14-01-66 [18.8K]
The answer is b
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8 0
3 years ago
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
4 years ago
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