Please answer I will give good rating and mark brainlest

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

4 years ago

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10

olchik [2.2K]

I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
$\epsilon = -N \frac{\Delta \Phi}{\Delta t}$
where
N is the number of turns in the coil
$\Delta \Phi$ is the variation of magnetic flux through the coil
$\Delta t$ is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
$\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb$
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
$\Phi_f = 0$

Therefore, we can calculate now the induced emf by using the first formula:
$\epsilon = -N \frac{\Delta \Phi}{\Delta t}=- (250) \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V$

7 0

4 years ago

If the planes of a crystal are 3.50 (1 A= 10^-10m = Ångstrom unit) apart, what wavelength of electromagnetic waves are needed so

goldenfox [79]

Answer:

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

Explanation:

We can use Bragg's Law's equation to solve this problem. The Bragg's Law's equation is written as follows:

mλ = 2d Sin θ

where,

m = order of reflection = 1

λ = wavelength = ?

d = distance between the planes of crystal = 3.5 x 10⁻¹⁰ m

θ = strike angle of waves on plane = 22°

Therefore, substituting the respective values in the equation, we get:

(1)λ = (2)(3.5 x 10⁻¹⁰ m)(Sin 22°)

4 0

3 years ago

Select all that apply. PLZ

Crank

A and c

do not complete the circuit there is no battery on a and paper dose not conduct electricity

5 0

3 years ago

19. What is shape of BF3

iren2701 [21]

Explanation:

The geometry of the BF 3 molecule is called trigonal planar (see Figure 5). The fluorine atoms are positioned at the vertices of an equilateral triangle. The F-B-F angle is 120° and all four atoms lie in the same plane.

4 0

3 years ago