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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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The speed of light in amber is

Umnica [9.8K]

Answer:

1.55

Explanation:

that it is the standard refractive index.

6 0

3 years ago

. You are making a frame of uniform width for a picture that is to be displayed at the local museum. The picture is 4 feet tall

nikdorinn [45]

Answer:

1 feet

Explanation:

Frame width = x

The length of the frame will be

$L=4+x+x$ (bottom and top that is why x is added twice)

$\\\Rightarrow L=4+2x$

The width of the frame will be

$W=3+x+x\\\Rightarrow W=3+2x$

Area is given by

$A=LW\\\Rightarrow 30=(4+2x)(3+2x)\\\Rightarrow 30=12+14x+4x^2\\\Rightarrow 4x^+14x-18=0$

Solving the above equation we get

$x=\frac{-14+\sqrt{14^2-4\cdot \:4\left(-18\right)}}{2\cdot \:4}, \frac{-14-\sqrt{14^2-4\cdot \:4\left(-18\right)}}{2\cdot \:4}\\\Rightarrow x=1, -\frac{9}{2}$

Hence width of the frame is 1 ft

4 0

3 years ago

When a mass is placed on a spring with a spring constant of 15 newtons per meter, the spring is compressed 0.25 meter. How much

Andre45 [30]

Answer:

0.47 J

Explanation:

The elastic potential energy of a spring is given as,

E = 1/2ke²........................ Equation 1

Where E = Elastic potential energy, k = spring constant, e = extension/compression.

Given: k = 15 N/m, e = 0.25 m.

Substitute into equation 1.

E = 1/2(15)(0.25)²

E = 0.46875

E ≈ 0.47 J.

Hence the elastic potential energy stored in the spring = 0.47 J

7 0

4 years ago

A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a

marshall27 [118]

Explanation:

The given data is as follows.

mass (m) = 0.160 kg, spring constant (k) = 8 n/m,

Maximum speed ( $v_{m}$ ) = 0.350 m/s

Formula for angular frequency is as follows.

$\omega = \sqrt{\frac{{k}{m}}$

$\omega = \sqrt{\frac{{8}{0.160}}$

$\omega$ = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

$\nu_{max} = A \omega$

A = $\frac{\nu}{\omega}$

= $\frac{0.35}{7.07}$

= 0.05 m

Hence, value of amplitude is 0.05 m.

(b) Displacement = 0.030 m

Formula for mechanical energy is as follows.

M.E = $\frac{1}{2}kA^{2}$

Putting the values into the above formula as follows.

M.E = $\frac{1}{2}kA^{2}$

= $\frac{1}{2} \times 8 \times (0.05)^{2}$

= $9.8 \times 10^{-3}$ Joule

For x = 0.03,

As, P.E = $\frac{1}{2} \times kx^{2}$

= $\frac{1}{2} \times 8 \times (0.03)$

= $3.6 \times 10^{-3}$

Hence, calculate the kinetic energy as follows.

K.E = M.E - P.E

= ( $9.8 \times 10^{-3}$ - $3.6 \times 10^{-3}$ ) J

= $6.2 \times 10^{-3}$ J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is $6.2 \times 10^{-3}$ J.

7 0

3 years ago

Can someone help me please :( ASAP. Also I’m sorry if it’s small

Umnica [9.8K]

Answer:

Graph #2

Explanation:

Notice, on the position time graph, there is a constant rate of change. This means that there is a constant velocity, meaning it's a straight line. The line is decreasing so the velocity will be negative.

3 0

3 years ago