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2 years ago

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A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

nd the other is at one end. What can be said about the forces exerted by the supports?

2 answers:

JulijaS [17]2 years ago

8 0

All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

gogolik [260]2 years ago

4 0

Answer:

The support at the middle of the board exerts an upward force that is equal to the weight of the board mg ( 24x9.81)N which is 235.44N.

The support at the other end exerts no force on the board since the board is already balanced by the middle support.

Explanation:

Detailed explanation and calculation is shown in the image below

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An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

Aneli [31]

We are given: Final velocity ( $v_f$ )=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

$v_f = v_i + at.$

Let us plug values of final velocity, and time in above equation.

$20=v_i+a(2.51)$

$20=v_i+2.51a$

Subtracting 2.51a from both sides, we get

$20-2.51a=v_i$ -----------equation(1)

Using another equation of motion

$v_f-v_i=2as$

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.

5 0

2 years ago

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

Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No

Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0

2 years ago

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea

Semmy [17]

Answer:

${\rm K} = 2.4\times 10^{-19}~J$

Explanation:

The electric field inside a parallel plate capacitor is

$E = \frac{Q}{2\epsilon_0 A}$

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

$F = qE = \frac{qQ}{2\epsilon_0 A}$

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

$C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60$

Plugging this identity into the force equation above gives

$F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q$

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is $1.6 \times 10^{-19}$

Therefore, the kinetic energy is $2.4\times 10^{-19}$

8 0

2 years ago

Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

$T^{2} = a^{3}$

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.

4 0

2 years ago