Ask question

Ask question

All categories

2 years ago

15

What is the strength of the electric field of a point charge of magnitude +4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A. 3.6

× 10^-4 N/C B. -2.7 × 10^-4 N/C C. -3.6 × 10^-4 N/C D. 2.7 × 10^-4 N/C

1 answer:

soldi70 [24.7K]2 years ago

8 0

Answer is D - from the equation of electric field for point charge

You might be interested in

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

b)

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

$E'=13 kt$ (13 kilotons)

which corresponds to

$E'=0.013 Mt$ (0.013 megatons)

Here the energy of the meteorite is equal to

$E=0.110 Mt$ (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

$\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46$

So, 8.46 bombs.

5 0

2 years ago

You are driving 12 meters per second. What is your speed in miles per hour? (1.6km=1 mike) EXPLAIN

Blizzard [7]

1.6km = 1 mike? Wow that guy is tall.

I think you meant 1.6 km = 1 mile? Okay then if we're going 12 meters per second how much would you travel in one hour? First we need to figure out how many seconds are in an hour. There are 60 seconds in a minute and 60 minutes in an hour so:

60×60=3600 seconds in an hour

Now we will multiply that by 12 meters per second and we get:

$12\frac{m}{s} \times 3600 s=43200m$

And 43200 meters is 43.2 km (1000 meters in 1 kilometer) meaning 43.2 kilometers an hour. Since there are 1.6 km in one mile we must divide 43.2km to 1.6.

$\frac{43.2}{1.6}=27mph$

And so your speed is 27 miles per hour.

8 0

3 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

2 years ago

What is the resistance of a nichrome wire at 0.0 ∘c if its resistance is 200.00 ω at 11.5 ∘c?

Fed [463]

The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relation between the two of them is:
<span>R = R0[1+ alpha(T-T0)] where:
</span>R is the new resistance we are looking for
alpha is the temperature coefficient of resistance. For nichrome, alpha = <span>4x10^-4/C
T0 is the standard temperature =11.5
R0 is the resistance at T0 = 200 ohms
T is the temperature at which we want to get R = 0

Substitute in the equation to get R as follows:
R = 200 [1+4*10^-4 (0-11.5)] = 199.08 ohm</span>

8 0

2 years ago

Which body part would most likely be affected by the improper application of a compression wrap?

Alborosie

It could be your arms or legs

4 0

3 years ago

Read 2 more answers