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docker41 [41]
3 years ago
11

Comic strip hero superman meets an asteroid in outer space and hurls it at 100m/s. he has a mass of "m". The astroid has a mass

of 1000m. Determine superman's velocity after throwing the astroid.
Physics
1 answer:
Katyanochek1 [597]3 years ago
8 0

Answer:

v = -10⁵ m/s

Explanation:

given,

speed of asteroid,v' = 100 m/s

mass of superman = m

mass of asteroid,M = 1000 m

recoil velocity of superman,v= ?

using conservation of momentum.

m u + M u' = m v + M v'

initial velocity of asteroid and superman is equal to  zero

 0 + 0 = m v + 1000 m x 100

m v = -100000 m

    v = -10⁵ m/s

superman's velocity after throwing the asteroid is equal to v = -10⁵ m/s

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A reconnaissance plane flies 545 km away from
Nikolay [14]

Answer:

681.6/ms

Explanation:

A reconnaissance plane flies 545 km away from  its base at 568 m/s. then flies back to its base  at 852 m/s.

What is its average speed?

Answer in its of m/s

Avg speed of the round trip is

2*568*852/(568+852)= 681.6/ms

6 0
3 years ago
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
Read 2 more answers
Josie sees lightning off in the distance. A few seconds later she hears thunder. What can Josie conclude?
Naddik [55]
Light wave travel faster than sound waves
8 0
3 years ago
Read 2 more answers
An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m
Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement  0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

     = \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0
3 years ago
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