Which of the following forces exists between objects even in the absence of direct physical contact

A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for

serg [3582]

Answer:

20.353125 V

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

q = Charge of proton = $1.6\times 10^{-19}\ C$

$v_A$ = Velocity of proton at point A = 50 km/s

$v_B$ = Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

$\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V$

The determined potential difference is 20.353125 V

3 0

3 months ago

Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma

Softa [3030]

The mass of the baked loaf will be lower than that of the dough.

7 0

2 months ago

Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} E ⃗ =(4000 j ^ +3000 k ^ ) N/

kicyunya [3294]

Answer:

Electric flux is calculated as $\phi=31562.63\ Nm^2/C$

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is $E=(4000j+3000k)\ N/C$

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

$\phi=E{\cdot}A$

$\phi=(4000j+3000k){\cdot}Ak$

Applying properties of the dot product, we calculate the electric flux as:

$\phi=3000\times Ak$

$\phi=3000\times \pi (1.83)^2$

$\phi=31562.63\ Nm^2/C$

Consequently, the electric flux for the circular area is $\phi=31562.63\ Nm^2/C$ . Thus, this represents the required answer.

4 0

3 months ago