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Viefleur [7K]
3 years ago
13

When practicing an oral presentation what can you do to prepare for the presentation

Physics
1 answer:
Igoryamba3 years ago
8 0
When practicing an oral presentation, you can prepare by writing a draft and practice reading aloud what you are going to say before your oral presentation.
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Technician A says that the device that controls shift points on an automatic transmission is the valve body. Technician B says t
alexira [117]

Answer: Technician B only

Explanation:

What actually controls the shifting in a automatic transmission is the vehicle's internal computer that helps shift the gears automatically. This includes a Control Module that adjusts the current provided to solenoids inside to control the gears The hydraulic system is used to select gears dependent on the transmission fluid's pressure applied.

3 0
3 years ago
RP COME AND GET IT!!!!!!!!
Ahat [919]
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2 years ago
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A 3 kg drone starts at 5 meter above the ground and reached 25 meter above the
docker41 [41]
Recall that work is the amount of energy transferred to an object when it experiences a displacement and is acted upon by an external force. It is given a symbol of W and is measured in joules (J).

W=\vec{F}\cdot \Delta \vec{d}

We can use this formula to determine the work done by very specific forces, generating specific types of energy. We will examine three types of energy in this activity: gravitational potential, kinetic, and thermal. Before we start deriving equations for gravitational potential energy and kinetic energy, we should note that since work is the transfer and/or transformation of energy, we can also write its symbol as \Delta E.
4 0
2 years ago
A 1.5-kilogram cart initially moves at 2.0 m/s. It is brought to rest by a constant net force in 0.3 seconds. What is the magnit
ser-zykov [4K]
We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s

Substitute their values, 
F = 1.5 * 6.67
F = 10 N

In short, Your Answer would be 10 Newtons

Hope this helps!
5 0
3 years ago
After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter
Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, \Delta L=3.1\ cm

(a) The change in length per unit original length is called strain. So,

\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193

(b) The force acting per unit area is called stress.

\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa

(c) The ratio of stress to the strain is called Young's modulus. So,

Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2

Hence, this is the required solution.

8 0
3 years ago
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