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ICE Princess25 [194]

4 years ago

8

The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's

velocity and acceleration are in opposite directions. Do you agree or disagree?

1 answer:

Ivan4 years ago

3 0

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity $v>0$ .

The acceleration $a$ is the rate of change of the velocity $v$ . This means that if $v$ is <em>increasing during</em> time, then $a$ must be positive. But if $v$ is <em>decreasing over</em> time, then $a$ will be negative (even though the velocity is positive).

Mathematically:

$a=\frac{dv}{dt}$

$v$ decreases ⇒ $\frac{dv}{dt}$

⇒ $a$ .

Example:

$v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}$

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0

4 years ago

Waves will have the highest speed in _____.air at 0°Cair at 5°Cair at 20°Cair at 30°C

Dahasolnce [82]

<span>The answer is air at 30oC . This is because air at 30oC has its molecules vibrate faster than at lower temperatures. Waves travel faster in molecules that are highly excited together than less excited. Sound may seem to travel further in cold air because the warm air above it refracts sound waves, at the boundary of the two air masses, back to the cold air hence its traveles greater distances. </span>

7 0

4 years ago

Read 2 more answers

If the mass of the Earth somehow increased with no change in radius, your weight would

Margaret [11]

Answer:

increase also

Explanation:

The weight of a person is equal to the gravitational pull exerted by the Earth on the person:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M$ is the mass of the Earth

$m$ is the mass of the person

$R$ is the Earth's radius

We notice that the weight is directly proportional to the mass of the Earth. Therefore, if the mass of the Earth M increases, and the radius R does not change, the weight of the person increases as well.

5 0

3 years ago

Read 2 more answers

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

aev [14]

Answer:

$x_2=1.60m$

Explanation:

From the Question We are told that

Initial Force $F_1=5800N$

Final Force $F_2=6500N$

Distance between the front and rear wheels \triangle x=3.20 m

Since

$\triangle x=3.20 m$

Therefore

$x_1+x_2=3.20$

$x_1=3.20-x_2$

Generally the equation for The center of mass is at x_2 is mathematically

given by

$x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}$

$x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}$

$2*F_1*x_2 =3.20F_1$

$x_2=1.60m$

6 0

3 years ago