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3 years ago

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Question 12 of 34 Which of the following means that an object is slowing down? /13 ယာ O A. Its velocity is decreasing. B. Its ac

celeration is increasing. O C. Its acceleration is decreasing. O D. Its velocity is increasing.

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

Answer:

A. Its velocity is decreasing

Explanation:

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This 80 kg car is moving at 20m/sec at the top where the hills radius is 100m. What is the centrifugal force?

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100 seconds is the right thing

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3 years ago

Swinging a golf club or baseball bat are examples of ______________ stretching.

kari74 [83]

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

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4 years ago

How do the meetings of the words exercise and fitness differ?

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3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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4 years ago

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

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4 years ago