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bezimeni [28]
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?
Physics
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

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kati45 [8]



Ask Siri or goggle has all answers



6 0
3 years ago
A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan
gladu [14]

Answer:

(a) L = 0·73 m

(b) 4·39 × 10^{-3} J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω =  (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>

Maximum potential energy =  (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy =  (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 × 10^{-3} J

∴ Maximum kinetic energy = 4·39 × 10^{-3} J

4 0
3 years ago
Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instant
Evgesh-ka [11]

Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5  ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

3 0
3 years ago
A hockey puck has a momentum of 3.8 kg•m/s [E]. If its
agasfer [191]

Answer:

1.6 \times 10 { }^{ - 1} kg \\

Explanation:

I hope, it helped you.

4 0
3 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
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