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3 years ago

Please help me , thank you

Physics

1 answer:

mash [69]3 years ago

5 0

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

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On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe

polet [3.4K]

The period of a simple pendulum is given by
$T= 2 \pi \sqrt{ \frac{L}{g} }$
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
$T_e= 2 \pi \sqrt{ \frac{L}{g_e} }$
where $g_e$ is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
$T_m= 2 \pi \sqrt{ \frac{L}{g_m} }$
where $g_m$ is the acceleration of gravity on the Moon.

If we do the ratio of the two periods, we get
$\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }$
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write $g_e = 6 g_m$ and we can rewrite the previous ratio as
$\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}$

so the period of the pendulum on the Moon is
$T_m = \sqrt{6} T_e = \sqrt{6} (1.6 s)=3.9 s$

8 0

4 years ago

When unbalanced force acts on a resting object the object what

garik1379 [7]

B. accelerates in the direction of the force. this is because the force is now unbalanced, causing the object to move in that direction. it's like pushing a pen off the table. you applied force in that direction and since the pen doesn't have enough inertia to withstand your unbalanced force, it rolls off the table.

5 0

4 years ago

If the ultimate shear strength of steel is taken to be 3.00 109 Pa, what force is required to punch through a steel plate 1.60 c

photoshop1234 [79]

Answer:

Explanation:

Given

Ultimate shear strength $\tau =3\times 10^9\ Pa$

thickness of steel Plate $t=1.60\ cm$

Cross-sectional $A=1.20\times 10^2\ cm^2$

suppose d is the diameter of cross-section

$\frac{\pi }{4}d^2=1.20\times 10^2$

$d=12.359\ cm$

Shear area of circular cross-section

$A_s=\pi dt$

$A_s=\pi \times 12.359\times 1.6$

$A_s=62.13\ cm^2$

Shear Force $F_s=A_s\times \tau$

$F_s=62.13\times 10^2\times 3000$

$F_s=18.639\ MN$

6 0

3 years ago

Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

statuscvo [17]

Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$