A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.

You start driving your car when the air temperature is 270.734 K. The air pressure in the tires is 454.518 kPa. After driving a

alisha [4.7K]

Answer:

T₂ = 280.17 K

Explanation:

Here, we can use the equation of state to find the final temperature of air in tires. Equation of State is written as follows:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Since, the volume is constant.

Therefore,

$V_{1} = V_{2} = V\\\\\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\T_{2} = \frac{T_{1}P_{2}}{P_{1}}$

where,

T₂ = Final Temperature of Air in Tire= ?

T₁ = Initial Temperature of Air = 270.734 K

P₁ = Initial Pressure of Air = 454.518 KPa

P₂ = Final Pressure of Air = 470.361 KPa

Therefore,

$T_{2} = \frac{(270.734\ K)(470.361\ KPa)}{454.518\ KPa}$

7 0

3 years ago

A constant force of 10 N is applied to an object. The object accelerates at a constant rate of 4 m/s2. What is the mass of the o

tatuchka [14]

Answer:

I think it is 2.5 kg.

Explanation:

I can say you do:

Fnet = m * a

So, you want to find mass:

m = Fnet / a

So, the answer you will receive is 2.5

7 0

3 years ago

A thin ball shell is shaped like a hollow ball with a radius of 42.8 cm. The material is a conductor with charge 0.8 nC A point

Gala2k [10]

Answer:

The Electric field will be 225.92 N/C

Explanation:

Given :

Radius of the hollow sphere R=42.8 cm
Initial charge on the conducting sphere is $Q_1=0.8\times10^{-9}\ \rm C$
Magnitude of the point charge $q_2=-5.4\times10^{-9}\ \rm C$

We know that the flux of the electric field through a close surface cab be calculated by using Gauss Law which is

$\int EdA=\dfrac{Q_{in}}{\epsilon_0}\\\\E\times4\pi\times 0.428^2=\dfrac{(0.8-5.4)\times10^{-9}}{\epsilon_0}\\\\E=225.92\ \rm N/C$

Hence the Electric Field is calculated.

The answer founded out by you seems to be correct and very close to the exact answer and the concept you have used in your answer is also correct.

5 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?

nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹ and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ = (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

ν = 3.394×10¹⁵ s-1

So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

4 0

4 years ago

A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.​

A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.