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rosijanka [135]
3 years ago
15

A stone falls freely from rest for 8.0s what is it final velocity

Physics
1 answer:
NemiM [27]3 years ago
7 0
Is that the full question?
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Holes drilled several kilometers into Earth’s crust provide direct evidence about Earth’s interior in the form of
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What would you do to improve the precision of an experiment?
GenaCL600 [577]

Explanation:

Precision represents that how close the different measurements of the sample one take are to one another.

  • One can increase the precision in lab by paying attention to each and every detail.
  • Usage of the equipment properly and also increasing the sample size.
  • Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
  • Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.
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3 years ago
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A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
tatyana61 [14]
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
7 0
3 years ago
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
3 years ago
A current-carrying wire 1.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10 A and there is
LenaWriter [7]

Answer:

0.2

Explanation:

F= BIL sin©

3= B×10×1.5 sin90

B=3/15

B= 0.2

8 0
2 years ago
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