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vodka [1.7K]
4 years ago
11

Friction always opposes an objects? A) Power B) Weight C) Motion D) Net force

Physics
2 answers:
marin [14]4 years ago
8 0
Friction always opposes motion
Feliz [49]4 years ago
7 0
Hey there Cndodson04p0v9cj,

Friction always opposes an objects _______

Answer:

Friction always opposes as objects (A) motion

Hope this helps :D

<em>~Natasha♥</em>
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A notebook is on a desk the force of the notebook weight pushing down on the table is 10 in the force of the table normal force
Mamont248 [21]

Answer:

Yes it is balanced

Explanation:

Because 10-10= a net force of 0N

Its equal (balanced)

5 0
3 years ago
You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to
goldenfox [79]

Answer:

f = 1.69*10^5 Hz

Explanation:

In order to calculate the frequency of the sinusoidal voltage, you use the following formula:

V_L=\omega iL=2\pi f i L         (1)

V_L: voltage = 12.0V

i: current  = 2.40mA = 2.40*10^-3 A

L: inductance = 4.70mH = 4.70*10^-3 H

f: frequency = ?

you solve the equation (1) for f and replace the values of the other parameters:

f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz      

The frequency of the sinusoidal voltage is f

3 0
3 years ago
The energy band gap of GaAs is 1.4eV. calculate the optimum wavelength of light for photovoltaic generation in a GaAs solar cell
Viktor [21]

Answer:

λ = 8.88 x 10⁻⁷ m = 888 nm

Explanation:

The energy band gap is given as:

Energy Gap = E = 1.4 eV

Converting this to Joules (J)

E = (1.4 eV)(1.6 x 10⁻¹⁹ J/1 eV)

E = 2.24 x 10⁻¹⁹ J

The energy required for photovoltaic generation is given as:

E = hc/λ

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = ?

Therefore,

2.24 x 10⁻¹⁹ J = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.24 x 10⁻¹⁹ J)

<u>λ = 8.88 x 10⁻⁷ m = 888 nm</u>

7 0
3 years ago
Any collection of things that have some influence on one another can be thought of as a system. When we talk about a system, we
Katen [24]

Answer:

B

Explanation:

Its B

6 0
4 years ago
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
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