All categories

1 month ago

How many electrons does it take to make 80 μc (microcoulombs) of charge?

1 answer:

6 0

The charge for a single electron is 1.602*10^ -19 C

80 µC can be expressed as 8*10^ - 5 C

This is basic arithmetic

Total Charge divided by the charge of one electron = Number of electrons

(8*10^ -5 C / 1.602*10^ -19 C) equals 4.99 * 10^14 electrons.

You might be interested in

A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of

serg [3582]

Provided Information:

Length of inclined plane = 8 m

Height of inclined plane = 2 m

Weight of the ice block = 300 N

Required Information:

Force needed to push ice block = F =?

Answer:

Force needed to push the ice block = 75 N

Explanation:

The required force to push this ice block up an inclined plane is given by

F = Wsinθ

where W is the weight of the ice block and θ is the angle indicated in the attached image.

Using trigonometric ratios,

sinθ = opposite/hypotenuse

where the opposite side is the height of the inclined plane and the hypotenuse is the length of the inclined plane.

Thus, sinθ = 2/8

θ = sin⁻¹(2/8)

which leads to θ = 14.48°

Therefore, F = 300*sin(14.48)

results in F = 75 N

This indicates that a force of 75 N is necessary to push the ice block on the specified inclined plane.

7 0

14 days ago

A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a

Softa [3030]

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

The weight component acting perpendicular to the slope, $mgcos \theta$ , directed into the slope
The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

The weight component aligned with the slope, $mgsin \theta$ , directed down the slope
The frictional force $F_f$ , acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

$a=0$

Thus, according to Newton's second law,

$F=ma$

implying the net force is zero:

$F=0$

Learn more about slopes and friction:

5 0

1 month ago

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

Keith_Richards [3271]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

$mg=m\frac{v^2}{r}$ where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top $g=9.8 m/s^2$

= loop radius

$v$ By solving for v,

$r=1200 m$

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from $v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

$v_A=550 km/h =152.8 m/s$

Given the plane's deceleration is consistent, we can obtain it using the following equation:

$s=\pi r=\pi(1200)=3770 m$

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$ - The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B: $N=m\frac{v_B^2}{r}$

$v_B$

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

6 0

1 month ago

Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass

serg [3582]

Utiliza Scoratic, funciona con cualquier tema.

5 0

1 month ago

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the

kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

In the case of the N-H-N combination:

The total force acting from this combination is:

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The force that thymine applies on adenine is:

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

When rounded to three significant figures, the net force is $8.86\times 10^{-9}N$

The negative value indicates that the force is attractive, as it is aimed towards thymi.

The force acting on the electron due to the proton is:

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

The ratio of the above forces is:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

5 0

1 month ago