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Musya8 [376]
3 years ago
10

if lightning is 3 times hotter than the sun and the sun can kill you instantly than why does a lighting strike not kill you inst

antly?
Physics
2 answers:
Mkey [24]3 years ago
5 0
A lightning strike happens so fast, and only hits a point of your body, but if you've noticed that it still completely melts the part of your body that it has stuck? Also the sun is a large mass of that heat, that will melts you the same way as lightning would if it were to continuously strike you all over your body. <span />
Ann [662]3 years ago
3 0
A lighting bolt just hit a single point in your body so it burn that part that it hits and leaves shockwaves on your skin, But if you were hit by the sun it would do the exact thing but not leave shockwave but just burn your hole skin and destroy everything because it not focus on just one point because the sun a huge compare to just a strike of lighting
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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
4 years ago
Describe two ways unbalanced forces help you in your day to<br> day life.
Alexxandr [17]

Answer:

  1. we need unbalance force to lift objects
  2. we need unbalance force to drag objects
6 0
3 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
3 years ago
The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre
frozen [14]

Answer:

a) P_m=1.78\ Pa

b) f=79.5775\ Hz

c) \lambda=7.076\ m

d) v=563.06\ m.s^{-1}

Explanation:

<u>Given equation of pressure variation:</u>

\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]

We have the standard equation of periodic oscillations:

\Delta P=P_m\ sin\ (kx-\omega.t)

<em>By comparing, we deduce:</em>

(a)

amplitude:

P_m=1.78\ Pa

(b)

angular frequency:

\omega=2\pi.f

2\pi.f=500

∴Frequency of oscillations:

f=\frac{500}{2\pi}

f=79.5775\ Hz

(c)

wavelength is given by:

\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{0.888}

\lambda=7.076\ m

(d)

Speed of the wave is gives by:

v=\frac{\omega}{k}

v=\frac{500}{0.888}

v=563.06\ m.s^{-1}

8 0
3 years ago
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e
Zinaida [17]

Answer:

F=1.38*10^{-9}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

Here k is the Coulomb constant. In this case, we have q_1=-e, q_2=e and d=4.09*10^-10m. Replacing the values:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

F=1.38*10^{-9}N

5 0
3 years ago
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