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1 year ago

Help?? ....... ┐(‘～`;)┌

Physics

2 answers:

7nadin3 [17]1 year ago

8 0

Answer:

(3) Both extensional as well as compressional strain is produced

Explanation:

Serggg [28]1 year ago

8 0

Both extensional as well as compressional strain is produced

Strain is amount of restoration formed by a restoring force .
It is formed in the direction of force applied.

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Can someone give me tips to passing AP Physics 1 with a solid A? Like, tips on how to memorize a lot of equations? Thank you!

Anna007 [38]

A good way for me to remember things is to study it, and to write it down! Say you want the formula for speed, I would write the formula multiple times on a piece of paper!

Here's a video that I haven't actually watched, I just looked it up! It might help you out though: <span>https://www.youtube.com/watch?v=-Wqrw4G79Kc</span>

3 0

2 years ago

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

3 years ago

Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horiz

steposvetlana [31]

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

$v_x = 3.80 m/s$

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

$v_y = u_y + gt$

where

$u_y =0$ is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

$v_y = 0 + (9.8)(1.00)=9.8 m/s$

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

$v=\sqrt{v_x^2 +v_y^2}=\sqrt{(3.8)^2+(9.8)^2}=10.5 m/s$

b) $\theta = -68.8^{\circ}$

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

$\theta = tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{-9.8}{3.8})=-68.8^{\circ}$

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so $v_x$ has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so $v_y$ is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

$\theta = -68.8^{\circ}$

4 0

3 years ago

Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. Straight upward incline with a speed of 2.95

Verdich [7]

Answer:

The length of the incline is 3.504 meters.

Explanation:

Let suppose that Julietta's ball decelerates uniformly, then we determine the length of the incline is determined by the following equation of motion:

$\Delta s = v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (Eq. 1)

Where:

$\Delta s$ - Length of the incline, measured in meters.

$v_{o}$ - Initial speed of the ball, measured in meters per second.

$a$ - Aceleration of the ball, measured in meters per square second.

$t$ - Time, measured in second.

If we know that $v_{o} = 2.95\,\frac{m}{s}$ , $t = 1.54\,s$ and $a = -0.876\,\frac{m}{s^{2}}$ , then the length of the incline is:

$\Delta s = \left(2.95\,\frac{m}{s} \right)\cdot (1.54\,s)+\frac{1}{2}\cdot \left(-0.876\,\frac{m}{s^{2}} \right) \cdot (1.54\,s)^{2}$

$\Delta s = 3.504\,m$

The length of the incline is 3.504 meters.

6 0

2 years ago

Two capacitors with capacitances of 3.25 and 5.00 μF, respectively, are connected in parallel. The system is connected to a 55-V

Nataliya [291]

The charge accumulated in 3.25 μF capacitor is 178.75 μC.

Answer:

Explanation:

In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.

As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.

C = Q/V

Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.

3.25× $10^{-6}$ = Q/55

Q = 3.25 * 55 = 178.75 μC

So the charge accumulated in 3.25 μF capacitor is 178.75 μC.

7 0

3 years ago

Help?? ....... ┐(‘～`;)┌​

Help?? ....... ┐(‘～`;)┌