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2 years ago

Torque can be calculated by multiplying the force

Physics

1 answer:

lana66690 [7]2 years ago

6 0

<span>Torque can be calculated by multiplying the force and the "perpendicular distance"

Hope this helps!</span>

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If you throw a baseball straight up,What is its velocity at the highest point?

sveta [45]

Answer: 52.

Explanation: Critical Thinking When a ball is thrown vertically up, it keeps going up until it hits a certain position, and then falls down. At the maximum point, the velocity is instantaneously zero.

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2 years ago

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A person drives for 1.00 h at 10.0 km/h and 2.00 h at 20.0 km/h. What is their average speed, in SI units?

Veronika [31]

I believe it is b if i’m correct

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2 years ago

Explain how wind erosion changes land forms

valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

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1 year ago

A 0.5 kg stone is raised from 1m to 2m height from the ground. what is the change in potential energy of the stone?

Usimov [2.4K]

Given: The mass of stone (m) = 0.5 kg

Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m

Acceleration due to gravity (g) = 9.8 m/s²

To find: The change in potential energy of the stone

Formula: The potential energy (P) = mgh

where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

Δ P = P₂ - P₁ = mg (h₂ - h₁)

or, = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)

or, = 4.9 J

Hence, the required change in the potential energy of the stone will be 4.9 J

4 0

2 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

2 years ago