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enyata [817]
3 years ago
9

A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.

Physics
1 answer:
earnstyle [38]3 years ago
4 0

Answer:

gfdvcfffddfgffffdrddgfddddghtscgvfrggxfhxdg

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Which of the following is true of our current knowledge of electrons?
frosja888 [35]
They have a negative charge and rotate around the nucleus
8 0
2 years ago
Zoe is shown two mystery boxes that are both 8,000cm3. Her teacher tells her that one mystery box is filled with rocks and the o
krek1111 [17]

Answer:

The box of rocks will have depression which can be seen without touching the box.

Explanation:

The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.

As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.

5 0
2 years ago
A. 2kg<br><br> B. 6kg<br><br> C. 8kg<br><br> D. 14kg
Sati [7]

Answer:

A

Explanation:

We know the force is (16N-12), which is 4N, and we know the acceleration is 2 m/s^2. Meaning, we can solve the formula m = F / a (mass equals force divided by acceleration), and we get 2kg.

7 0
2 years ago
Both cars start from the same point. Which describes the motion shown?
evablogger [386]
Is there a graph we can look at?
4 0
3 years ago
Read 2 more answers
The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm
DIA [1.3K]

Answer:

T_{max} = 4.735\,kN\cdot m

Explanation:

The shear stress due to torque can be calculed by using the following model:

\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}

The maximum torque on the section is:

T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}

The Torsion Constant for the circular tube is:

J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})

J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]

J_{tube} = 4.560\times 10^{-6}\,m^{4}

Now, the require output is computed:

T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}

T_{max} = 4.735\,kN\cdot m

7 0
3 years ago
Read 2 more answers
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