A Mars rover has a mass of 955 kg. What is its weight in Newtons on Mars?

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

3 years ago

Dropping a rock from 4 feet in the air is an example of what type of energy?

Gre4nikov [31]

Answer:

The type of energy is potential energy

3 0

3 years ago

Read 2 more answers

Una carga de 4 uC penetra perpendicularmente en un campo magnetico de 0.4 T con una velocidad de 7.5x10 4 m/s. Calcular la fuerz

tiny-mole [99]

Answer:

F_B = 0.12N

Explanation:

In order to calculate the magnetic force on the charge, you use the following formula:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

q: charge of the particle = 4μC = 4*10^-6 C

v: speed of the charge = 7.5*10^4 m/s

B: magnitude of the magnetic field = 0.4T

The direction of the motion of the charge is perpendicular to the direction of the magnetic field. Then, the magnitude of the magnetic force is:

$F_B=qvBsin90\°\\\\F_B=(4*10^{-6}C)(7.5*10^4 m/s)(0.4T)=0.12N$

The magnetic force on the charge is 0.12N

3 0

3 years ago

What is the electrical current passing through this ammeter?

Arturiano [62]

The pointer on the meter's scale is pointing to 8.5 .

3 0

3 years ago

A student makes a homemade resistor from a graphite pencil 5.00 cm long, where the graphite is 0.05 mm in diameter. The resistiv

Alex

Answer:

(a) RC time constant of the circuit is 6.9 × 10⁻⁶ ms

(b) The potential drop across the capacitor 1.00 s after the switch is closed is 0 V

Explanation:

The given parameters are

The length of the graphite pencil, L = 5.00 cm

The diameter of the graphite, D = 0.05 mm

The resistivity of the graphite, ρ = 1.38 × 10⁻⁵ Ω/m

The capacitance of the capacitor, C = 10.0 mF

The voltage of the power source, V = 0.50-V

(a) The RC time constant of the circuit, τ, is given as follows;

τ = R × C

Where;

R = The resistance of the graphite = L × ρ

C = The capacitance of the capacitor

∴ R = 5.00 cm × 1.38 × 10⁻⁵ Ω/m = 6.9 × 10⁻⁷ Ω

RC time constant of the circuit, τ = 6.9 × 10⁻⁷ Ω × 10.0 mF = 6.9 × 10⁻⁶ ms

RC time constant of the circuit, τ = 6.9 × 10⁻⁶ ms

(b) The potential drop after t = 1.00 s is given as follows;

$i = \dfrac{V}{R} \cdot e^{-\dfrac{t}{R\cdot C} }$

Where;

I = The current in the circuit

V = The voltage in the circuit = 0.50 V

R = resistance in the circuit = 6.9 × 10⁻⁷ Ω

C = The series capacitance = 10.0 mF

t = The time taken = 1.00 s

Plugging in the variable values, gives;

$I = \dfrac{0.5}{6.9 \times 10^{-7}} \cdot e^{-\dfrac{1.00}{6.9 \times 10^{-7}\times 10.0 \ mF} } = 0$

V(1) = I·R = 0 × R = 0

The potential drop across the capacitor 1.00 s after the switch is closed, V(1) = 0 V

7 0

3 years ago