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3 years ago

Which body could NOT be called a gas giant?

2 answers:

6 0

Well, considering we are speaking of the planets in our solar system. There are 8 (9 if you like pluto
Not gas giants: mercury, Venus, earth, mars, Pluto.

Gas giants: Saturn, Jupiter, Uranus, Neptune.

Dafna11 [192]3 years ago

4 0

Sorry I don’t know but good luck and hope you find it

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Consider N non-interacting diatomic molecules stuck on a metal surface. Each molecule can either lie flat on the surface, in whi

Art [367]

Answer:

Detailed solution given in diagram

7 0

3 years ago

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

Problem 5 A block of mass 3 kg slides on a horizontal, rough surface towards a spring with k = 500 N/m. The kinetic friction coe

vredina [299]

Answer:

Explanation:

given,

mass of block = 3 kg

spring constant k = 500 N/m

kinetic friction coefficient µk = 0.6

speed of block = 5 m/s

F = µk N

F = 0.6 x 3 x 9.8

F = 17.64 N

using energy conservation

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2+Fx$

$\dfrac{1}{2}\times 3 \times 5^2=\dfrac{1}{2}\times 500 \times x^2+17.64\times x$

250 x² + 17.64 x - 37.5 = 0

on solving

x = 0.354 m

graph is attached below

4 0

4 years ago

Read 2 more answers

To the nearest tenth, what is the area of the shaded segment when JA=8ft ??

Murrr4er [49]

$A=\dfrac{1}{6}\pi 8^2-\dfrac{8^2\sqrt{3}}{4}\approx 5.8\;[ft^2]$
Answer: <span>C. 5.8 ft squared</span>

7 0

3 years ago

Read 2 more answers

An airplane cruises at 850 km/h relative to the air. It is flying from Denver, Colorado, due west to Reno, Nevada, a distance of

skad [1K]

Answer:

difference in flight time= 0.3023 hour

Explanation:

The question is incomplete, but I found it in your textbook.

Spped of aircraft = 850 km/h

Opposing speed of wind = 90km/h

Hence, the net speed when it's travelling west = 850 - 90 = 760 km/hr

The distance covered = 1200km

time taken = distance/ time = 1200/ 760 = 1.5789 hours

When coming back, the speed of the wind is complementary to the speed of the aircraft so

net speed when it's coming back = 850 +90 = 940 km/hr

time taken in this instance = 1200/ 940 = 1.2765 hours

Hence, the difference in flight time= 1.5789 - 1.2765 = 0.3023 hour

8 0

3 years ago