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3 years ago

A light microscope allows for more magnification than electron microscope because it uses a beam of visible light. true false

Physics

1 answer:

Igoryamba3 years ago

4 0

False because light microscopes have low resolve and magnification.

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A proton, mass 1.67 × 10−27 kg and charge +1.6 × 10−19 c, moves in a circular orbit perpendicular to a uniform magnetic field of

stellarik [79]

Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s

Explanation:

For proton, the centripetal force required for circular motion is provided by the magnetic force,

so Fm= Fc

q v B = m v²/r

m= mass of charged particle

v= velocity

B =magnetic field

q= charge

r= radius of circular path

v= q B r/m

now v= r ω

ω= angular velocity

ω r = q B r /m

ω=q B /m

now ω= 2π/T where T =time period

so 2π/T=q B/m

T= 2 πm/q B

T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]

T= 7.28 x 10⁻⁸ s

6 0

3 years ago

What is heat energy!?

nordsb [41]

Answer:

heat, energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred—i.e., heat flows—from the hotter body to the colder. example: stove

Explanation:

hope this helps

5 0

2 years ago

Read 2 more answers

PLS Help

ale4655 [162]

Answer:

sport is an activity that involves movement and teamwork.

hobby is a personal interest that you enjoy doing often.

Explanation:

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2 years ago

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$