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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?

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