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2 years ago

The idea that people behave in odd ways on a full moon is an example of

Physics

2 answers:

garik1379 [7]2 years ago

8 0

Old wives' tales, superstition, bubbe meisses, urban legend, folk myth, rumor, pseudoscience.

dimulka [17.4K]2 years ago

6 0

Superstition i think.....

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Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp

Studentka2010 [4]

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

$v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}$

Since it is given that

$v_{a}=0.6c\\\\v_{b}=0.6c$

Applying values in the formula we get

$v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c$

8 0

2 years ago

Two ways that weak nuclear forces and electromagnetic forces are simaliar

STALIN [3.7K]

Have very short ranges

6 0

2 years ago

A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

6 0

2 years ago

Does the diagram show a spring or neap tide?How do you know?

vampirchik [111]

It shows a neap tide

6 0

2 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

2 years ago