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Nady [450]
3 years ago
7

In a lesson about the behavior of gases, Genaris and her classmates learn that the volume of a gas is affected by its temperatur

e. They formulated a hypothesis: If the temperature of a gas increases, the volume of the gas will increase because the increased temperature will increase the kinetic energy of the particles, which will then result in the expansion of the gas. The results of their experiment supported the hypothesis. They wanted to know if there were other factors that affect the volume of a gas. What would Genaris and her classmates most likely do to test this idea? They would formulate a new hypothesis with the same independent variable but a different dependent variable as the original hypothesis. They would formulate a new hypothesis with a different independent and dependent variable as the original hypothesis. They would formulate a new hypothesis with the same dependent variable but a different independent variable as the original hypothesis. They would formulate a new hypothesis that has the independent and dependent variables of the original hypothesis switched.
Physics
2 answers:
Liula [17]3 years ago
7 0
<span>To know if there were other factors that affected the volume of a gas, Genaris and her classmates should: </span>"formulate a new hypothesis with the same dependent variable but a different independent variable as the original hypothesis." In this case, the dependent variable is the volume of the gas and the new independent variable is a factor they think will affect the volume of the gas.
ss7ja [257]3 years ago
7 0

The correct answer is

B). They would formulate a new hypothesis with the same dependent variable but a different independent variable as the original hypothesis.

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When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is
tangare [24]

Answer:opposite

Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed

6 0
3 years ago
How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t
Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

7 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
Describe how the design of a vacuum flask keeps the liquid inside hot ?
JulijaS [17]
The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection 
3 0
3 years ago
Read 2 more answers
A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs
adell [148]

Answer:

the magnitude of the charge Q on each plate is 3.053 *10^{-8} \ C

Explanation:

Given that :

mass (m) = 7.00 *10 ^{-3} \ kg

charge (q) = +0.155 µC = +0.155 *10^{-6}\ C

angle \theta = 30^0 \ C

Area A on each plate = 0.0135 m²

From the diagram below;

tan \ \theta = \frac{Eq}{mg}    ----- equation (1)

Also by using Gauss Law ;

Q = \epsilon_0 \phi

Q = \epsilon_0EA     ----- equation (2)

Combination equation 1 and 2 together ; we have

Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}

Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}

Q = 3.053 *10^{-8} \ C

8 0
3 years ago
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