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4 years ago

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What is one advantage a primary source has over a secondary source? A. A primary source always includes several perspectives. B.

A primary source combines several different accounts of an event. C. A primary source offers a firsthand account of an event. D. A primary source generally benefits from having been written after an event.

1 answer:

Ostrovityanka [42]4 years ago

5 0

Answer: a primary source I typically free of outside Interpretation

Explanation:

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A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola

irakobra [83]

C I’m pretty sure!!!!!

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3 years ago

In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit

Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

$x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}$

Putting, t = 1 s and x = 4 m in above equation, we get :

$4 = \dfrac{a(1)^2}{2}\\\\a = 8 \ m/s^2$

Therefore, the acceleration of the cart is 8 m/s².

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3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

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

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

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

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

3 years ago

Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou

ch4aika [34]

Answer:

2s

Explanation:

Given parameters:

Distance = 30m

Speed = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

Speed = $\frac{disance }{time}$

Time taken = $\frac{distance }{speed }$ = $\frac{30}{15}$ = 2s

The time it takes is 2s

6 0

3 years ago

Read 2 more answers