All categories

3 years ago

A railroad car containing an angry bull is standing onthe

Physics

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

$v_{f}$ = - M / m v

Explanation:

We must define a system formed by the wagon and the torro, in this case the forces of the movement are internal, so the moment is preserved, write the moment in two moments

Initial. Before the bull's movement, in this case the two are still

p₀ = 0

Final. The bull is moving

$p_{f}$ = M v + m $v_{f}$

p₀ = $p_{f}$

0 = M v + m $v_{f}$

We cleared the train speed

$v_{f}$ = - M / m v

The negative sign indicates that the train moves in the opposite direction of the bull

You might be interested in

Which is the poorest conductor of energy

sdas [7]

Insulators are the poorest conductor of energy

7 0

4 years ago

Read 2 more answers

Is this wave transverse or longitudinal

lys-0071 [83]

It depends on if it is a sound wave or light wave

8 0

3 years ago

Problem:

kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a = -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

7 0

3 years ago

A rotating wheel requires 2.90-s to rotate through 37.0 revolutions. Its angular speed at the end of the 2.90-s interval is 97.2

goldenfox [79]

Answer:

Angular acceleration, $\alpha =20.32\ rad/s^2$

Explanation:

It is given that,

Displacement of the rotating wheel, $\theta=37\ rev=232.47\ radian$

Time taken, t = 2.9 s

Initial speed of the wheel, $\omega_i=0$

Final speed of the wheel, $\omega_f=97.2\ rad/s$

Let $\alpha$ is the angular acceleration of the wheel. Using the third equation of kinematics to find it as :

$\alpha=\dfrac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha=\dfrac{(97.2)^2}{2\times 232.47}$

$\alpha =20.32\ rad/s^2$

So, the angular acceleration of the wheel is $20.32\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago

In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

= (10970)²/(2x220)

= 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

7 0

3 years ago