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4 years ago

Explain how the velocity of an object changes in respect to time.

Physics

2 answers:

Zepler [3.9K]4 years ago

8 0

Answer:

The rate of change in the velocity of an object per unit time is referred as acceleration and the kind of motion is known as accelerated motion. The acceleration is positive when it is in the direction of velocity and negative when it is opposite to the direction of velocity.

Explanation:

HOPE YOU UNDERSTAND

blondinia [14]4 years ago

4 0

Answer:

Is this your ans of this question

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Select the generic equation suggested by a graph showing a half-parabola. a y = kx b y = kx2 c y = k(1/x)

guajiro [1.7K]

Answer:

its b i took the test

Explanation:

7 0

2 years ago

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A middle-aged man typically has poorer hearing than a middle-aged woman. In one case a woman can just begin to hear a musical to

Dmitriy789 [7]

Answer:

Explanation:

$I_0$ = Threshold intensity for woman

I = Intensity of sound for man

$\beta$ = Intensity level of sound = 6 dB

Intensity level of sound is given by

$\beta=10log\dfrac{I}{I_0}\\\Rightarrow 6=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{6}{10}=log\dfrac{I}{I_0}\\\Rightarrow 10^{\dfrac{6}{10}}=\dfrac{I}{I_0}\\\Rightarrow \dfrac{I}{I_0}=10^{\dfrac{6}{10}}\\\Rightarrow \dfrac{I}{I_0}=10^{0.6}\\\Rightarrow \dfrac{I}{I_0}=3.98\approx 4$

The ratio is 4

8 0

4 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiati

QveST [7]

Answer:

8.67×10⁻⁶ N/m²

Explanation:

p = Momentum of a photon

E = Energy of a photon

c = Speed of light

I = Intensity of light

Force = dp/dt

$p=\frac{E}{c}$

$\\\Rightarrow F=\frac{\frac{dE}{dt}}{c}$

As given in question

$F=2\frac{\frac{dE}{dt}}{c}$

Now F/A = Pressure

$P=\frac{2I}{c}\\\Rightarrow P=\frac{2\times 1300}{3\times 10^{8}}\\\Rightarrow P=8.67\times 10^{-6}\ N/m^2$

∴ Magnitude of the pressure on the sail is 8.67×10⁻⁶ N/m²

4 0

4 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim

never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

$\text{Distance}=20\times 0.5=10\ m$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m$

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

$\text{Distance}=u\times 0.5=0.5u\ m$

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0$

$u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s$

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

5 0

3 years ago

Read 2 more answers