By examining a wave's pattern after it interacts with a barrier or gap, measurements can be made to better understand certain wa

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

serg [3582]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

$e=-1.6\cdot 10^{-19}C$

$q_2 = Ne$

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

$e=-1.6\cdot 10^{-19}C$

$q_1 = -Ne$

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>

6 0

3 months ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

serg [3582]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We are tasked with finding the pressure here

Applying the pressure formula

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v refers to velocity

Insert the values into the equation

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Therefore, the pressure at this moment is 0.875 mPa

5 0

4 months ago

A trained sea lion slides from rest down a long

Keith_Richards [3271]

Answer:

1.5 m/s²

Explanation:

Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.

Considering the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

This gives us N = mg cos θ

Next, examining the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substituting for N yields:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

hence a = g (sin θ − μ cos θ)

If we set θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

this results in a = 1.48

When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².

5 0

2 months ago

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

ValentinkaMS [3465]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, $F_1 = 32\ N$ (to the right)

The force exerted by Simone, $F_2 = -16\ N$ (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

6 0

3 months ago

Read 2 more answers

On a foggy day, you are driving at 70 mph and Granny is driving at 50 mph. As you try to pass her, you both see an overturned tr

inna [3103]

Response:

C?

Reasoning:

7 0

3 months ago