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My name is Ann [436]
3 years ago
13

A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do

you think the energy required will be greater? Why?
Physics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

Specific heat at constant pressure is =  1.005 kJ/kg.K

Specific heat at constant volume is =  0.718 kJ/kg.K

Explanation:

given data

temperature T1 =  50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

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A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

7 0
3 years ago
A scientist classifies some plants into two groups
gizmo_the_mogwai [7]

Answer:

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3 years ago
A parachutist weighs 1000N. when she opens her parachute, it pulls upwards on her with a force of 2000N. (a) draw a diagram to s
Nataly_w [17]

(a) Find free-body diagram in attachment (please rotate the picture, such that R points upward and mg points downward)

There are only two forces acting on the parachutist:

- Its weight, downward, of magnitude 1000 N, labelled with "mg" in the diagram (where m = mass of the parachutist, g = acceleration of gravity)

- The air resistance, upward, of magnitude 2000 N, labelled with "R" in the diagram

As the air resistance is larger than the weight, in the diagram it is represented with a longer arrow, in order to show the difference in magnitude.

(b) 1000 N, upward

By taking upward as positive direction, we can rewrite the two forces as:

R = +2000 N

mg = -1000 N

Where we have written the weight as a negative number, since its direction is downward.

Therefore, the net force on the parachutist will be

F = R + mg = +2000 + (-1000) = +1000 N

And the positive sign indicates that the resultant force is upward.

(c) The parachutist will be accelerated upward (= he/she will slow down)

We can answer this part by applying Newton's second law, which is summarized by the following:

F=ma (1)

where

F is the resultant force on the body

m is its mass

a is its acceleration

For the parachutist in this problem, the mass is

m=\frac{mg}{g}=\frac{1000 N}{9.8 N/kg}=102 kg

Therefore, using (1), we find the acceleration of the parachutist:

a=\frac{F}{m}=\frac{+1000}{102}=+9.8 m/s^2

Where the positive sign indicates that the acceleration is upward. Therefore, the parachutist will be accelerated upward, which means that he/she will slow down, since its direction of motion was downward.

4 0
3 years ago
Please help! Science and force stuff. (picture included)
Nikitich [7]
#4. It says "someone sitting in a chair" So the objects involved are the person and the chair. D is the answer because it talks about the action/reaction forces between the person and the chair.

#5. Work = Force (parallel to direction of distance) * distance. A, B and D are not correct because there is no distance.  C. is the best answer
4 0
3 years ago
A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe
Alchen [17]
A)
2revs in 0.08s
   so in 1s thats 25revs
therefore thats <u>50π radians</u> in one second

b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
<u>v=47.123ms^-1</u>

c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
<u>f=25Hz</u>

6 0
3 years ago
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