An aircraft acceleration from 100m/s to 300m/s in 100 s what is acceleration

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

2 years ago

A racecar driver on a flat racetrack steps on the gas, changing her velocity from 10 m/sec to 30 m/sec in 5 seconds. What is the

erastova [34]

Answer:

4m/s2

Explanation:

The following data were obtained from the question:

U (initial velocity) = 10m/s

V (final velocity) = 30m/s

t (time) = 5secs

a (acceleration) =?

Acceleration is the rate of change of velocity with time. It is represented mathematically as:

a = (V - U)/t

Now, with this equation i.e

a = (V - U)/t, we can calculate the acceleration of the race car as follow:

a = (V - U)/t

a = (30 - 10)/5

a = 20/5

a = 4m/s2

Therefore, the acceleration of the race car is 4m/s2

8 0

2 years ago

Read 2 more answers

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:

love history [14]

Answer:

Rate of dissolving compounds:

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

Boiling Point of Compounds:

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0

2 years ago

Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?

earnstyle [38]

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

6 0

2 years ago

An aircraft acceleration from 100m/s to 300m/s in 100 s what is acceleration​

An aircraft acceleration from 100m/s to 300m/s in 100 s what is acceleration