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3 years ago

⚠Heeeeeelelepelepleplepelepeplepe⚠

Physics

1 answer:

snow_tiger [21]3 years ago

4 0

Answer:

It's B, anything about a circle is Stationary

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7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-

netineya [11]

Answer:

Torque = 882Nm

Explanation:

Torque = Mg×distance

But plank's is pivoted ,therefore distance=3/2=1.5m

Mass of Nancy=60jg

Acceleration due to gravity, g=9.8m/s^2

Torque= 60×9.8×1.5

Torque= 882Nm

6 0

3 years ago

Which stage follows the red giant stage of star development, if the star is a high-mass star?

Anestetic [448]

I believe it's a supernova

3 0

4 years ago

I need help with this answer

7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0

3 years ago

A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field

HACTEHA [7]

Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

$v=\frac{E}{B}$

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

$v=\frac{0.5}{2.3}=0.217 m/s$

3 0

3 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

8 0

3 years ago