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Nat2105 [25]
2 years ago
13

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

yringe tightly. 3. Pull the plunger back to double the volume of gas in the syringe. Which best describes the purpose and outcome of the demonstration? This is a demonstration of Charles’s law. As the volume increases, the temperature decreases, and the marshmallow will freeze. This is a demonstration of Charles’s law. As the volume increases, the temperature increases, and the marshmallow will melt. This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. This is a demonstration of Boyle’s law. As the volume increases, the pressure increases, and the marshmallow will shrink.
Physics
2 answers:
lana [24]2 years ago
7 0
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger. 
</span><span>
 Keisha follows the instructions for a demonstration on gas laws.
 1. Place a small marshmallow in a large plastic syringe.
 2. Cap the syringe tightly.
 3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change.  Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
MrRissso [65]2 years ago
5 0

Answer:

C) This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

Explanation:Took the quiz on edg

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A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w
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Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

As the proton is coming at rest after travelling 0.200 m to the right,  vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C

⇒ E = 1.04*10⁻¹ N/C

5 0
2 years ago
5. A wave with peaks separated by .34 m has a wavelength of ________________m.
Eduardwww [97]

Answer: 0.17 I think

Explanation:

I asked a doctor

8 0
2 years ago
A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const
hoa [83]

Answer:

4 (please see the attached file)

Explanation:

While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.

So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.

When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.

This acceleration is related with the tangential acceleration, by this expression:

at = α*r

This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.

7 0
2 years ago
a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)
rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)

Now, using the given information, we have:

mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}

Calculating, we have:

N_{x}=mg*Sin(\alpha)

N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N

Hence, we have that the x-component of the normal force is equal to  <u>1706.45 N.</u>

Have a nice day!

3 0
2 years ago
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Anton [14]

Answer:

Explanation:

moment of inertia of each blade which is similar to rod rotating about its one end

= 1/3 ml²

moment of inertia of 3 blades = ml²

= 5500 x 46²

I = 11638 x 10³ kg m²

angular velocity = 2πn where n is rotation per second

n = 11 / 60

angular velocity = 2π x 11/60

= 1.1513 rad /s

angular momentum

= moment of inertia x angular velocity

=  11638 x 10³ x 1.1513

= 13399 x 10³ kg m² per second.

4 0
2 years ago
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