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3 years ago

Total potential and kinetic energy of an object.

Physics

1 answer:

Ne4ueva [31]3 years ago

8 0

Answer:

The Total Mechanical Energy

As already mentioned, the mechanical energy of an object can be the result of its motion (i.e., kinetic energy) and/or the result of its stored energy of position (i.e., potential energy). The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

Explanation:

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snow_tiger [21]

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3 years ago

Which sequence of events in emotional responses is characteristic of the james-lange theory of emotion?

AlladinOne [14]

According to James-Lange theory of emotion, a stimulus first leads to bodily arousal, and this is followed by our interpretation of it as an emotion.

4 0

3 years ago

Which one of the following is a derived SI unit?<br>A.newton B.meter C.mole d.Kilogram

marusya05 [52]

Answer:

Meter

Explanation:

I'd say meters, cause it's the SI unit of length,

which is a Derived Quantity.

5 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

2 years ago