What force is necessary to accelerate a 50kg box at 30 m/s2?

DENIUS [597]

No because sugar is made up of organic material

5 0

2 years ago

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do

ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

So

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

$C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K$

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

$C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\$

So

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

8 0

3 years ago

As the wavelength increases, the frequency (2 points) Select one:

LiRa [457]

A.
According to the equation: v=λ*f
λ stands for wavelength. wavelength increases, the frequency will decrease. lower frequency deserves low energy.

7 0

3 years ago