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2 years ago

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"An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 19.0 k

g,19.0 kg, a radius of 0.260 m,0.260 m, and a length of 0.650 m.0.650 m. The mass of the end of the barrel equals a fifth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is ????

1 answer:

Sophie [7]2 years ago

5 0

Answer:

the acceleration due to gravity, g of the cylinder is 9.81m/s2

Explanation:

since the cylinder is rolling down from a hill, the acceleration due to gravity of any object coming down from the atmosphere is always constant which is g=9.8m/s2 or approximately 10m/s2

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In a series circuit, a generator (1300 Hz, 12.0 V) is connected to a 14.0- resistor, a 4.40-μF capacitor, and a 6.00-mH inductor

klemol [59]

Answer:

(a) 2.8 V

(b) 5.6 V

(c) 9.8 V

Explanation:

Given:

Frequency of the generator (f) = 1300 Hz)

Terminal voltage (V) =12.0 V

Resistance of resistor (R) = 14.0 Ω

Capacitance of capacitor (C) = 4.40 μF = 4.40 × 10⁻⁶ F

Inductance of the inductor (L) = 6.00 mH = 6.00 × 10⁻³ H

In order to find the voltages across each, we first need to find the reactance and impedance.

Reactance of the inductor is given as:

$X_L=2\pi f L\\\\X_L=2\times 3.14\times 1300\times 6.00\times 10^{-3}\\\\X_L=49\ \Omega$

Reactance of the capacitor is given as:

$X_C=\frac{1}{2\pi fC}\\\\X_C=\frac{1}{2\times 3.14\times 1300\times 4.40\times 10^{-6}}\\\\X_C =28\ \Omega$

Now, impedance is given as:

$Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega$

Current across the circuit is given as:

$I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A$

As resistor, capacitor and inductor are connected in series, the current across each of them is same and equal to total current in the circuit.

(a)

Voltage across the resistor is given as:

$V_R=IR\\\\V_R=0.2\times 14=2.8\ V$

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

$V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V$

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

$V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V$

Therefore, the voltage across the inductor is 9.8 V.

6 0

2 years ago

An object has a mass of 15 kg and is accelerating to the right at 16.3 m/s2. The free-body diagram shows the horizontal forces a

marshall27 [118]

Refer to the free body diagram shown melow.

F = applied force
R = frictional force
m = 15 kg, the mass of the object

The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N

The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.

Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)

Answer:
The most reasonable answer is R = 5.5 N

8 0

2 years ago

Read 2 more answers

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0

2 years ago

In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wear

Ivahew [28]

Answer: $P=14.6 W$

Explanation:

According to the Stefan-Boltzmann law for real radiating bodies:

$P=\sigma A \epsilon T^{4}$ (1)

Where:

$P$ is the energy radiated (in Watts)

$\sigma=5.67(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A$ is the Surface area of the body

$T=30\°C + 273.15= 303.15 K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin

$\epsilon=0.6$ is the body's emissivity

On the other hand, we are told the human body is roughly approximated to a cylinder of length $L=2.0m$ and circumference $C=0.8m$ .

The circumference of a circle is: $C=0.8m=2 \pi r$ where $r$ is the radius. Hence $r=\frac{0.8m}{2 \pi}=0.1273 m$ .

Now we have to input this value for $r$ in the Area of a cylinder formula:

$A=\pi r^{2}L$

$A=\pi (0.1273 m)^{2}(2 m)$

$A=0.0509 m^{2}$ (2)

Substituting (2) in (1):

$P=(5.67(10)^{-8}\frac{W}{m^{2} K^{4}}) (0.0509 m^{2}) (0.6) (303.15 K)^{4}$ (3)

Finally:

$P=14.62 W \approx 14.6 W$

7 0

2 years ago

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin

fgiga [73]

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of $(\vec{E}\times\vec{B})$ vector is the direction of propagation of the wave .

Direction of magnetic field = $\hat{j}$

$B=B_{0}\sin(kx-\omega t)\hat{k}$

We need to calculate the poynting vector

Using formula of poynting

$\vec{S}=\dfrac{E\times B}{\mu_{0}}$

Put the value into the formula

$\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}$

$\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

Hence, The poynting vector is $\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

7 0

2 years ago