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Natalka [10]
3 years ago
12

To make extra money during the summer, Mr. Garber, a 66.0-kg physics teacher, paints the outside of houses while sitting on a 5.

0-kg plank suspended by three vertical cables. What is the tension in each of the three cables?
Physics
1 answer:
deff fn [24]3 years ago
6 0
The total mass of Mr. Garber and the plank is 71kg. We can find the tension in the cable by dividing the gravitational force by 3. So let's solve for the gravitational force.

F_{g} = (71kg)*(9.81m/s) = 696.51N

Next divide by three because the tension is split.

F_{T} = 696.51/3 = 232.17N

F_{T} = 232.17N
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ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

v_v^2 + v_h^2 = v^2

v_v^2 + 13^2 = 14^2

v_v^2 = 196 - 169 = 27

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When the skier jumps to the its potential energy is converted to kinetic energy:

E_p = E_k

mgh = mv_v^2/2

where m is the skier mass and h is the vertical distance traveled, v_v is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

gh = v_v^2/2

h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m

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3 years ago
How do oxygen and beryllium atoms transform into oxygen ion O2- and Be2 beryllium ion Be2? tysm
Hoochie [10]
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Katen [24]

Answer:

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Explanation:

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as in the exercise indicate that the velocities perpendicular to the magnetic field,

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         F = m a

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let's calculate

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The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

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\vec F = I (\vec L \times  \vec B)

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