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3 years ago

a gas exerts less pressure when it has a a. smaller volume b. lower temperature c. higher temperature d. two of the above

1 answer:

6 0

The combined gas law is
$\frac{Pressure_{1} *Volume_{1}}{Temperature_{1}} = \frac{Pressure_{2} *Volume_{2}}{Temperature_{2}}$ with 1 and 2 both being anything with pressure, volume, and temperature. Since pressure is on the top and temperature is on the bottom, they are inversely related, meaning that when the temperature gets high the pressure goes low. In addition, since pressure and volume are both on the top, when volume goes down pressure goes down too. Therefore, since A and C are right, D is the correct answer

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In one type of solar energy system, sunlight heats the air within solar panels, which heats copper tubes filled with water. What

Anna71 [15]

The type of energy that is produced by the system that is described is heat energy. The correct answer is D.

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3 years ago

Read 2 more answers

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

Nitella [24]

Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0= 20² + (2 x a x 240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

3 0

2 years ago

In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy

myrzilka [38]

<h2>Answer:</h2>

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = $\frac{1}{2}kx^2$ --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = $\frac{1}{2}k(4)^2$

E = 8k

<em>Make k subject of the formula</em>

k = $\frac{E}{8}$ [measured in J/cm]

<em>From the second statement;</em>

<em>It is stretched by 4cm.</em>

This means that total elongation will be 4cm + 4cm = 8cm.

The potential energy stored will be found by substituting the value of x = 8cm and k = $\frac{E}{8}$ into equation (i) as follows;

U = $\frac{1}{2}\frac{E}{8} (8)^2$

U = $\frac{1}{2}{8E}$

U = ${4E}$

Therefore, the potential energy stored will now be 4 times the original one.

3 0

3 years ago

- Una sustancia posee un punto de fusión de -110ºC y un punto de ebullición de 6ºC. Explica en qué estado físico se encontrarán

Burka [1]

Answer:

B

Explaination:

...

3 0

3 years ago

A proton, moving with a velocity of vii⁄, collides elastically with another proton that is initially at rest. Assuming that the

KATRIN_1 [288]

Answer:

Explanation:

Given that,

Mass of proton

Mp=m=1.627×10^-27kg

One of the proton is at rest then it velocity is 0

Let the other proton be moving at vi

Since the after collision the two proton moves together with the same velocity(i.e inelastic collision)

Then, using conservation of energy

Kinetic energy before collision = kinetic energy after collision

Given that, K.E=½mv²

Before collision = after collision

½mvi²+½m(0)²=½(m+m)vf²

½mvi²=½(2m)vf²

½mvi²=mvf²

Divide through by m

½vi²=vf²

vi²=2vf²

Take square root of both sides

√vi²=√(2vf²).

vi=√2 ×vf

Then, the final velocity is

vf = vi /√2

b. Direction of the velocity vectors after collision

Let vf1 be the final velocity for the incident proton

vf2 be the final velocity for the proton initially at rest.

Conserving momentum in the ˆj direction

Piy = 0 = Pfy = mp•vf1y +mp•vf2y

vf1y = −vf2y

So the protons have equal magnitude speeds in the ˆj direction. Because the speed of the particles are equal, the magnitude of their speeds in the ˆi direction should also be equal |vf1x| = |vf2x|.

Conserving momentum in the ˆi direction.

Pix = mp•vi = Pfx = mp•vf1x +mp•vf2x = 2mp•vfx

Then, vfx =vi/2

Using the Pythagorean theorem to solve for the magnitude of vfy

vf²=vi²/2= vfx² +vfy² =vi²/4+vfy²

vfy = vi√(½-¼)=vi/2=vfx

So because the ˆi and ˆj components of vf are the same, both protons are deflected away at an angle of θ = 45° from the ˆi direction, with opposite ˆj components (so the angle between vf1 and vf2 is 90°).

3 0

3 years ago