I need this to be correct and explane why its that answer.

A donkey starts from rest and accelerates for 6.7 s at a rate of 1.6 m/s².

vitfil [10]

The maximum speed of the donkey is 10.72m/s

The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.

It is given that donkey attains an acceleration of 1.6 m/s^2

The time taken to accelerate to given speed is 6.7 seconds

We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

a is the acceleration of the donkey

t is the time taken in seconds

v = u + at

v= 0 + 1.6 x 6.7

= 10.72 m/s

Hence the donkey obtains the speed of 10.72 m/s

For further reference:

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3 0

1 year ago

An example of an atom that has no charge is one that has?

Pavlova-9 [17]

Answer: Equal number of protons and electrons. Example: an atom of oxygen atom has 8 electrons and 8 protons and is neutral.

Explanation:

An atom that has no charge is a neutral atom. It contains electrons equal to protons. For example: A neutral atom of oxygen has 8 protons and 8 electrons.

An atom which has charge is said be ionized. It is either positively charged or negatively charged. It is positively charge when the number of electrons is less than the number of protons. For example: $Na^+$ contains 10 electrons and 11 protons.

And when the number of electrons is greater than the number of protons, the atom is negatively charged. For example, $Cl^-$ has 17 protons and 18 electrons. It readily accepts an electron to complete its octet.

3 0

2 years ago

Read 2 more answers

A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac

lana66690 [7]

Answer:

charges of the beads is 1.173 × $10^{-15}$ C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know force for mass

force = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 × $10^{9}$ )

so charges of the beads = 1.173 × $10^{-15}$ C

3 0

2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw

aalyn [17]

Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons

4 0

2 years ago