A measurement must include both a number and a(an)

A car is stopped at a traffic light. When the light turns green, the car starts from rest with an acceleration of 2.5 m/s2 . Als

erma4kov [3.2K]

Answer:

a) f = 453.3Hz

b) f = 443.1Hz

c) f = 420Hz

Explanation:

First of all we need to know the positions and velocities of both vehicles.

For the car:

Xc(-5) = 0m; Vc(-5) = 0m/s

$Xc(5) = 2.5*5^2/2=31.25m$ ; Vc(5)=2.5*5 = 12.5m/s

$Xc(10)=2.5*10^2/2=125m$ ; Vc(10)=2.5*10=25m/s

For the truck:

Xt(-5)=10*(-5) = -50m; Vt(5)=10m/s

Xt(5)=10*5 = 50m; Vt(5)=10m/s

Xt(10)=10*10=100m; Vt(10)=10m/s

Now for part a) t=-5s. The truck is behind the car, so:

$f=\frac{C}{C-Vt}*fo = 453.3Hz$

Now for part b) t=5s. The car is behind the truck, so:

$f=\frac{C+Vc}{C+Vt}*fo = 443.1Hz$

Now for part b) t=10s. The truck is behind the car, so:

$f=\frac{C-Vc}{C-Vt}*fo = 420Hz$

8 0

3 years ago

Please help me now.

aev [14]

I think it is option 2

8 0

3 years ago

If we decrease the size of a quantum dot that contains an electron, what happens to the energy levels?

Law Incorporation [45]

We will solve this problem using the direct concept related to band gap energy, that is, a band gap is the distance between the valence band of electrons and the conduction band, i. e, the energy range in a solid where no electron states (Electronic state) can exist Mathematically can be described as,

$E_n = \frac{h^2n^2}{8mcR^2}$

Where,

h = Planck's constant

n = Energy level

mc = Effective mass of the point charge

R = Size of the particle

As you can see the energy is inversely proportional to the size of the particle:

$E_n \propto \frac{1}{R^2}$

Therefore if the size is decreased, the amount of energy is increased.

6 0

2 years ago

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

2 years ago

Which of the following would NOT produce a current? A. Moving a bar magnet inside a coil of wire O B. Moving a bar magnet toward

sineoko [7]

Answer:

c

Explanation:

7 0

2 years ago