What is a suggested solution to a scientific problem

Will mark brainliest

son4ous [18]

Answer:

One 200 is bigger than 100

Explanation:

6 0

3 years ago

Read 2 more answers

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w

Yuki888 [10]

Explanation:

Formula for the electric field due to the infinite sheet of charge is as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

where, $\sigma$ = surface charge density

Now, formula for electric force acting on the proton is as follows.

F = eE

where, e = charge of the proton

According to the Newton's second law of motion, the net force acting on the proton is as follows.

F = ma

a = $\frac{eE}{m}$

= $\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}$

= $\frac{e \sigma}{2m \epsilon_{o}}$

According to the kinematic equation, speed of the proton in perpendicular direction is as follows.

$v_{f} = v_{i} + at$

= $(0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t$

= $\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}$

= 683.974 m/s

Hence, total speed of the proton is as follows.

v' = $\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}$

= $\sqrt{921600 + 467820.43}$

= $\sqrt{1389420.43}$

= 1178.73 m/s

Therefore, we can conclude that speed of the proton is 1178.73 m/s.

3 0

3 years ago

A unit for measuring frequency is? ohm watt ampere hertz

pishuonlain [190]

"Frequency" is a measure of how frequent something is.

Its unit is the "Hertz".

1 Hertz = 1 per second

8 0

3 years ago

How does absorption spectra occur?

Arlecino [84]

An absorption spectrum occurs when light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.

hope this helped :)

mark brainliest

3 0

4 years ago

A convex lens has a focal length of 0.33 m. The object distance is 0.7 m. What is the image distance?

dimulka [17.4K]

Answer:

Explanation:

1/v - 1/u = 1/f

given, f = 0.3 m, u = -0.4m

so, 1/v - 1/-0.4 = 1/0.3

or, 1/v = 1/0.3 - 1/0.4 = 1/1.2

v = 1.2 m

now, differentiating 1/v - 1/u = 1/f with respect to t,

-1/v² dv/dt + 1/u² du/dt = 0

or, dv/dt = (v/u)² du/dt

putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m

so, dv/dt = (1.2/-0.4)² × 0.01

= 0.09 m/s

hence, speed of image with respect to lens is 0.09 m/s .

from formula of magnification

magnification, m = v/u

differentiating with respect to time both sides,

dm/dt = (u dv/dt - vdu/dt)/u²

= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²

= (-0.036 - 0.012)/0.16

= -0.048/0.16

= -0.3 m/s

hence, magnitude of rate of change of lateral magnification is 0.3 m/s

7 0

3 years ago