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3 years ago

Cumulative Exam

Physics

2 answers:

Pani-rosa [81]3 years ago

8 0

Answer:

A. Plane mirrors produce only virtual images, and concave mirrors produce real and virtual images.

Explanation:

dalvyx [7]3 years ago

4 0

Answer:

Explanation:

Hi! It would be A. Plane mirrors produce only virtual images,and concave mirrors produce real and virtual images.

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Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther

photoshop1234 [79]

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

8 0

3 years ago

T=2pi square root 1/g solve for g. Explanation would be really helpful.

Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

7 0

3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, radial ( $a_{r}$ ) and tangential ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is: