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2 years ago

What is true weightlessness ?

Physics

1 answer:

alukav5142 [94]2 years ago

3 0

Answer:

weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.

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An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Rudik [331]

The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ $\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π $0.25^{2}$
A1=π $0.50^{2}$

we get. R2 = 16R

7 0

3 years ago

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

$T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}$

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

Thus, the new period will be reduced by 50%

8 0

3 years ago

michael kicks a ball at an angle if 36* horizontal. its initial velocity is 46 m/s. Find the maximum height it can reach, total

MAVERICK [17]

(a) At its maximum height, the ball's vertical velocity is 0. Recall that

${v_y}^2-{v_{0y}}^2=2a_y\Delta y$

Then at the maximum height $\Delta y=y_{\mathrm{max}}$ , we have

$-\left(\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=37\,\mathrm m$

(b) The time the ball spends in the air is twice the time it takes for the ball to reach its maximum height. The ball's vertical velocity is

$v_y=v_{0y}+a_yt$

and at its maximum height, $v_y=0$ so that

$0=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\sin36^\circ+\left(-9.8\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$\implies t=2.8\,\mathrm s$

which would mean the ball spends a total of about 5.6 seconds in the air.

$x=v_{0x}t$

so that after 5.6 seconds, it will have traversed a displacement of

$x=\left(46\,\dfrac{\mathrm m}{\mathrm s}\right)\cos36^\circ(5.6\,\mathrm s)$

$\implies x=180\,\mathrm m$

3 0

2 years ago

4. What's the voltage drop if V, = 56 V, V2 = 35 V, and V3 = 3 V

Murrr4er [49]

Answer:

The correct option is A. 18 V

Therefore,

$V_{1}=18\ V$

Explanation:

Given:

Total Voltage,

V = 56 V,

Voltage drop

$V_{2}=35\ V$

$V_{3}=3\ V$

To Find:

$V_{1}=?$ (Assumed ,as it is not given clearly in the question)

Solution:

So in series combination total Voltage is given as

$V=V_{1}+V_{2}+V_{3}$

Substituting the values we get

$56=V_{1}+35+3\\\\V_{1}=56-38=18\ V$

Therefore,

$V_{1}=18\ V$

5 0

3 years ago

"A jack-in-the-box is a toy in which a figure in an open box is pushed down, compressing a spring. The lid of the box is then cl

Svetlanka [38]

We know, Applied force(f) = Spring constant (K) * Extension of material (x)
Here, f = 12 N
x = 0.070 m

Substitute their values,
12 = k * 0.070
k = 12 / 0.070
k = 171.43 N/m

In short, Your Answer would be 171.43 N/m

Hope this helps!

7 0

3 years ago

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What is true weightlessness ?​

What is true weightlessness ?