Which color of the visible light waves has the highest energy level?

If we want to describe work , we must have what ?

ivanzaharov [21]

An external force that is being applied in the direction of the displacement

4 0

3 years ago

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago

How can we use thermal energy to our advantage?

devlian [24]

Answer:

One of the primary advantages of thermal power is that the generation costs are extremely low. No fuel is needed to generate the power, and the minimal energy needed to pump water to the Earth's surface can be taken from the total energy yield.

Explanation:

8 0

3 years ago

Read 2 more answers

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago

A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelera

user100 [1]

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

Vf₁ = 25 m/s

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

s₂ = 90 m

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

5 0

3 years ago