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atroni [7]
3 years ago
15

For an isothermal process, the work done by or on a system of ideal gas is equal to the change in what?

Physics
2 answers:
miss Akunina [59]3 years ago
8 0
I think the answer is B
Jlenok [28]3 years ago
4 0
T<span>he work done is equal to the change in heat energy</span>
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if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet
klio [65]

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

Where:

Q  is the Volume flow rate

r=23 cm \frac{1 m}{100 cm}=0.23 m  is the effective radius

L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

\Delta P=1.88(10)^{4} Pa  is the difference in pressure

\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

4 0
4 years ago
If tje object is placed pf the focus of a concave mirror the image is formed....​
gavmur [86]

Answer:

with the mirror into the right direction

Explanation:

4 0
3 years ago
Read 2 more answers
Josh starts throwing his pitch; we assume that his arm moves in a perfect circular arc, and his arm is 0.7 m long. If he needs t
harina [27]

As we know that centripetal acceleration is given as

a_c = \frac{v^2}{R}

here we know that

linear speed = 35 m/s

here radius of path is same as length of arm

R = 0.7 m

now from above above formula we have

a_c = \frac{35^2}{0.7}

a_c = 1750 m/s^2

so acceleration is 1750 m/s/s

5 0
4 years ago
A constant net force that has a magnitude of 135.5 N is exerted on a 26.7 kg object that is initially not moving. a. Draw a free
jek_recluse [69]

Explanation:

Given that,

Net force = 135.5 N

Mass = 26.7 kg

(a). We need to draw a free body diagram

A force is exerted on a object.

(b). We need to calculate the acceleration

Using formula of acceleration

a =\dfrac{F}{m}

a=\dfrac{135.5}{26.7}

a=5.07\ m/s^2

(c). We need to draw the sketch a position vs. time graph for the object.

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times5.07\times t^2

s=2.535 t^2

(d). We need to draw the sketch a velocity vs. time graph for the object.

Using equation of motion

v = u+at

Put the value into the formula

v=0+5.07t

v =5.07 t

(e). We need to calculate the distance travel in 8.82 s

Using equation of motion

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}\times5.07\times(8.82)^2

s=197.20\ m

Hence, This is the required solution.

8 0
3 years ago
use the general formulas for gravitational force and centripetal force to derive the relationship between speed and orbital radi
MrRissso [65]

Answer:

v = \sqrt{\frac{GM}{r}}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{Mm}{r^{2}}  (1)                

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

The centripetal force can be found by means of Newton's second law:

F = ma  (2)

Since it is a circular motion, the acceleration can be defined as:

a = \frac{v^{2}}{r}  (3)

Where v is the velocity and r is the orbital radius.

Replacing equation (3) in equation (2) it is gotten:

F = m\frac{v^{2}}{r}  (4)

Hence,

m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}

Then, v can be isolated:

mv^{2} = G\frac{Mmr}{r^{2}}

mv^{2} = G\frac{Mm}{r}

v^{2} = G\frac{Mm}{mr}

v^{2} = \frac{GM}{r}

v = \sqrt{\frac{GM}{r}}

So the relationship between speed and orbital radius is given by the expression v = \sqrt{\frac{GM}{r}}

8 0
3 years ago
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