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2 years ago

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What is Vatiable Velocity?

1 answer:

Allushta [10]2 years ago

7 0

Answer:

Explanation: When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity. Example: A rotating fan at a constant speed has variable velocity, because of continuous change in direction.

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anyone know where I can find stuff (answer key, tables, etc.) for my Newton's Law of Motion lab report on edge2020? need answers

Zolol [24]

Usually the full tables are found on quizlet or quizzes

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2 years ago

What The BEST GAME?????/<br><br> a: Roblox<br><br> b; Mincraft<br><br> c; Fortnite

Maksim231197 [3]

Answer:

A,C

Explanation:

8 0

3 years ago

Read 2 more answers

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag

VARVARA [1.3K]

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$ . Hence, this is the required solution.

5 0

3 years ago

Determine the value of the resultant and its location from O.<br>see attach image.

xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L

the magnitude of the resultant will be

F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]

(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L

(√⅛)[d] = 5L/24

d = 5L/24 / (√⅛)

d = 5√⅛L/3

8 0

2 years ago

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago