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Natalija [7]
3 years ago
11

In a butcher shop, a horizontal steel bar of mass 4.94 kg and length 1.46 m is supported by two vertical wires attached to its e

nds. The butcher hangs a sausage of mass 2.49 kg from a hook that is at a distance of 0.10 m from the left end of the bar. What is the tension in the right wire?

Physics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

Tension in right wire = 25.9N

Explanation:

I have attached a free body diagram to depict this question.

From the diagram, i have labelled the tensions in the strings T1 and T2.

While i labelled the weight of the bar as Wb and weight of sausage as Ws.

Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)

Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.

Thus;

T1 + T2 = Wb + Ws

We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N

Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N

Thus,

T1 + T2 = 48.46N + 24.43

T1 + T2 = 72.89N - - - - - (eq 1)

Now, let's take moments about the left end of the bar.

The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m

So, moments about left end;

T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)

1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)

1.46T2 = 35.373 + 2.443

1.46T2 = 37.816

T2 = 37.816/1.46 = 25.9N

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uranmaximum [27]

The number of revolutions the electron completes in 60.0-μs of the strike is 134.

A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

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7 0
2 years ago
the royal Gorge Bridge in Colorado rises 321 m above the Arkansas river. suppose you kick a rock horizontally off the bridge. Th
KengaRu [80]

Answer:

2.48 m/s

Explanation:

We can use the kinematic equation,

s = ut +½at²

Where

s = displacement

u = initial velocity

t = time taken

a = acceleration

Using the equation in vertical direction,

321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0

We get t = 8.01 s

Using the equation in the horizontal direction,

52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction

So u = 2.48 m/s

5 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
If a river current is 8.0 m/s, and a boat is traveling 10.0 m/s upstream, what is the boat’s speed relative to the riverbank?
Norma-Jean [14]
If the boat is i travling at 10 m/s and the river is 8.0 m/s the boats speed is 18.0 m/s

3 0
3 years ago
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