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podryga [215]
1 year ago
5

Many gates at railway crossings are operated manually. A typical gate consists of a rod usually made of iron, consisting heavy w

eight. What is the need for the heavy weight at one end of the rod??

Physics
1 answer:
alekssr [168]1 year ago
7 0

Answer:

  Making the crossing gate balanced makes the mechanical drive simpler and less expensive to operate.

Explanation:

See the attachment for a picture of what we're talking about.

The gate must be rotated up and down. That rotation is easier if the mass being rotated is balanced about the center of rotation. Less torque (and work) is required to create the desired motion.

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Which of the following statements describes an interaction between the geosphere and atmosphere?
german
B is the answer, I’m really good at this subject
8 0
1 year ago
Read 2 more answers
(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is u
stiv31 [10]

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = 3.95\times 10^{- 25}\ kg

Mass of Uranium- 238, m' = 3.90\times 10^{- 25}\ kg

Velocity, v = 3.00\times 10^{5}\ m/s

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

The radius of the ion in a magnetic field is given by:

R = \frac{mv}{qB}

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})

\Delta x = 2(\frac{m - m'}{qB}v)

Now,

By putting suitable values in the above eqn:

\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm

\Delta x = 1.25\ cm

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

3 0
1 year ago
A clever inventor has created a device that can launch water balloons with an initial speed of 85.0 m/s. Her goal is to pass a b
seropon [69]

Answer:

She should launch the balloon at an angle of 59.9° above the horizontal.

Explanation:

Please, see the attached figure for a graphical description of the problem.

The position and velocity vectors of the water balloon at time "t" can be obtained using the following equations:

r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)

v = (v0 · cos θ, v0 · sin θ + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

θ = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity vector at time "t".

Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.

At the maximum height (276 m), the vector velocity of the balloon is horizontal (see v1 in the figure). That means that the y-component of the velocity vector is 0. Then, using the equation of the y-component of the velocity vector, we can write:

At maximum height:

vy = v0 · sin θ + g · t

0 = v0 · sin θ + g · t

We also know that at maximum height, the y-component of the position vector is 276 m (see r1y in the figure). Then:

At maximum height:

y = y0 + v0 · t · sin θ + 1/2 · g · t²  

276 m = y0 + v0 · t · sin θ + 1/2 · g · t²

So, we have two equations with two unknowns (θ and t):

276 m = y0 + v0 · t · sin θ + 1/2 · g · t²

0 = v0 · sin θ + g · t

To solve the system of equations, let´s take the equation of the y-component of the velocity and solve it for sin θ. Then, we will replace sin θ in the equation of the y-component of the position to obtain the time and finally obtain θ:

0 = v0 · sin θ + g · t

0 = 85.0 m/s · sin θ - 9.81 m/s² · t

9.81 m/s² · t / 85.0 m/s = sin θ

Replacing sin θ in the equation of the vertical component of the position:

276 m = y0 + v0 · t · sin θ + 1/2 · g · t²    (y0 = 0)

276 m = 85.0 m/s · t · (9.81 m/s² · t /85. 0 m/s) - 1/2 · 9.81 m/s² · t²

276 m = 9.81 m/s² · t² - 1/2 · 9.81 m/s² · t²

276 m = 1/2 · 9.81 m/s² · t²

276 m / ( 1/2 · 9.81 m/s²) = t²

t = 7.50 s

Now, we can calculate the angle θ using the equation obtained above:

9.81 m/s² · t / 85.0 m/s = sin θ

9.81 m/s² · 7.50 s / 85.0 m/s = sin θ

θ = 59.9°

She should launch the balloon at an angle of 59.9° above the horizontal.

6 0
1 year ago
Three masses are placed along the x-axis: m1 = 3.2 kg is a distance x1 = 0.43 m to the left of the origin, m2 = 3.6 kg is a dist
faust18 [17]

Answer:

0.153 meters to the right of the origin.

Explanation:

The location of the center of mass of a particle system is given by this formula:

\overline x = \frac{\Sigma_{i = 1}^n (m_{i} \cdot x_{i})}{\Sigma_{i = 1}^n m_{i}}

The current inputs are presented below (It is assumed that distances are positive for all number to the right of the origin:

x_{1} = - 0.43 m, m_{1} = 3.2 kg\\x_{2} = + 0.31 m, m_{2} = 3.6 kg\\x_{3} = + 0.47 m, m_{3} = 4.1 kg\\

Then, the location is finally found:

\overline x = \frac{(-0.43 m)\cdot (3.2 kg)+(0.31 m)\cdot (3.6 kg) + (0.47 m)\cdot (4.1 kg)}{3.2 kg + 3.6 kg + 4.1 kg}

\overline x = + 0.153 m

5 0
2 years ago
Find the direction of the sum of<br> these two vectors:<br> 3.14 m,<br> 30.0°<br> 60.0°<br> 2.71 m
Gemiola [76]

Answer:

60.0⁰ is the answer problem solution

5 0
1 year ago
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