Car 1 of mass m1 is waiting at a traffic light.

Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown

9966 [12]

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

4 0

3 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

Learn more about capacitors:

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#LearnwithBrainly

5 0

3 years ago

At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re

densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ $-( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}$

where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

$v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s$

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

$-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}$