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Jobisdone [24]
3 years ago
13

SO what exactly is beyond our atmosphere, like space, has anybody actually been in space ????? or are we just making hopeless fa

ntasies
Physics
1 answer:
Mkey [24]3 years ago
8 0

Answer:

The universe, milky Way even black holes. Yes the first people who went to the moon put a flag to represent America. We are not making hopeless fantasies.

Explanation:

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A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.
Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

v_f = \dfrac{872.10-851.10}{2}

v_f =10.5\ Hz

v_o = 872.1 - 10.5

V_0 = 861.6\ Hz

The speed of jogger

v = \dfrac{v_1 \times 343}{v_0}-343

v = \dfrac{872.1 \times 343}{861.6}-343

v = 4.18 m/s

5 0
3 years ago
Is water wet my friends have been arguing about this for about 3 months now but now it is time to hear other people
nirvana33 [79]
Water is not wet, wet is a state of being. Water, however, can make things wet.
8 0
3 years ago
Read 2 more answers
A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th
BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

s = ut + \frac{1}{2}at^2

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

156.3 = 31\times t + 0

t = \frac{156.3}{31 }

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 m/s^2

Substituting in h = ut + \frac{1}{2}gt^2

h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2

h = 124.694 m

So height of ramp = 124.694 m

3 0
3 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
A boat cruises 60 meters west in 10 seconds. What is its instantaneous velocity?
kari74 [83]

A - 6 meters a second ( 6m/s)

3 0
3 years ago
Read 2 more answers
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