Ask question

Ask question

All categories

3 years ago

13

A comic-strip superhero meets an asteroid in outer space and hurls it at 100 m/s. The asteroid is a thousand times more massive

than the superhero is. In the strip, the superhero is seen at rest after the throw. Taking physics into account, what would be his recoil speed? What is this in miles per hour?

1 answer:

Molodets [167]3 years ago

5 0

Answer:

100000 m/s in the opposite direction

22369.4185194 mph

Explanation:

$m_1$ = Mass of superhero

$m_2$ = Mass of asteroid = $1000m_1$

$v_2$ = Speed at which the asteroid is thrown = 100 m/s

$v_1$ = Recoil velocity

Here the linear momentum of the system is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=-\dfrac{1000m_1\times 100}{m_1}\\\Rightarrow v_1=-100000\ m/s$

The recoil speed is 100000 m/s in the opposite direction

Converting to miles per hour

$\dfrac{10000}{1609.34}\times 3600=22369.4185194\ mph$

The speed is 22369.4185194 mph

You might be interested in

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

If you were to move to the Canadian North Woods, what adaptations or behavioral changes would you make?

iragen [17]

Adaptation will mean taking action to minimize the negative effects of change. ... the use of new tools and techniques for decision-making, For example, projected increases in drought, fire, windstorms, and insect and disease outbreaks are expected to result in greater tree mortality. Fewer trees will reduce Canada’s timber supply, which in turn will affect the economic competitiveness of Canada’s forest industry. This would leave forestry-dependent communities vulnerable to job losses, closure of forestry processing facilities and an overall economic slump.

5 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi

Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.

energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ

Sometimes they use 10 instead of 9.8 for the g constant.

Rember to make me Brainliest!!!

3 0

3 years ago

HELP ASAP!!!! SCIENCE LOVERS !!! What do the lines making a circle around the LOW-PRESSURE area indicate?

Degger [83]

1 is D.
2 is A.

hope i helped

5 0

3 years ago

Read 2 more answers