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3 years ago

A student increases the temperature of a 300 cm^3 balloon from 30c to 60c. what will the new volume of the ballon be

2 answers:

5 0

<span>This problem uses the ideal gas law, PV = nRT. T for temperature must be in Kelvin, so we add 273.15 to the Celsius temperatures. We are then changing T from 303.15 K to 333.15K. The ratio of the temperatures 333.15/303.15 ~= 1.1. So holding pressure constant, the new volume of the balloon will be approximately 10% larger, or 330cm^3.</span>

Mama L [17]3 years ago

4 0

From Charles law the volume of a fixed mass of gas is directly proportional to the absolute temperature when pressure is kept constant.
Volume α temperature, mathematically; k = V/T
Therefore, comparing two gases V1/T1 = V2/T2
V1 =300 cm³ T1= 30°C (30 +273= 303K) V2= ? T2= 60°C (60 +273= 333K)
Thus, 300/303 = V2/333
V2 = (300 × 333)/ 303
= 329.703 cm³
Hence the new volume will be 329.703 cm³

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m \cdot g \cdot \sin \theta$

$F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}$

$F = 78.606\,N$

b) The equations of equilibrium are the following:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m\cdot (a + g \cdot \sin \theta)$

$F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})$

$F = 88.911\,N$

3 0

3 years ago

Newtons law of garvitation is known as universal law. Why

xeze [42]

Answer:

It is called so because it is applicable on all bodies having mass, and the bodies will be governed by the same law, that is newton's law of gravitation. Thus, as it is applicable universally, it is called as universal law.

5 0

3 years ago

Who is known as the father of India

balandron [24]

Answer:

the answer is

Mahatma Gandhi

6 0

3 years ago

Read 2 more answers

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

Find the tension in the two ropes that are holding the 4.2 kg object in place.

Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0

3 years ago