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skelet666 [1.2K]

3 years ago

9

If you traveled one mile at a speed of 100 miles per hour and another mile at a speed if 1 mile per hour, your average speed wou

ld not be (100mph+1 mph)/2 or 50.5. What you’d be your average speed?
physic killing my brain cell

1 answer:

Semenov [28]3 years ago

7 0

Answer:

v = 1.98 mph

Explanation:

Given that,

Speed to travel one mile is 100 mph

Speed to travel another mile is 1 mph

The formula used to find your average speed is given by :

$v=\dfrac{2v_1v_2}{v_1+v_2}$

Putting the values, we get :

$v=\dfrac{2\times 100\times 1}{100+1}$

v = 1.98 mph

So, yours average speed is 1.98 mph.

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Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as

malfutka [58]

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2 $\pi \sqrt{0.5*9.8*0.3}$

T=1.2042s

The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

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6 0

1 year ago

A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for

luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

$\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}$

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

$\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}$

In conclusion, the normal force acting on the block is 50 N.

5 0

2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

When an object reflects all the light waves that strike it looks white or black?

egoroff_w [7]

Light waves can be from any color, depends on what it is bouncing no or reflecting off of.

3 0

3 years ago