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TEA [102]
2 years ago
10

Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble

A was initially moving at a velocity of 0.5 m/s, but after the collision, it has a velocity of −0.1 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.
Physics
1 answer:
stich3 [128]2 years ago
3 0

From the calculations, the final momentum of B is 8.16 m/s

<h3>What is conservation of linear momentum?</h3>

According to the principle of the conservation of linear momentum, the momentum before collision is equal to the total momentum after collision.

This implies that;

MaUa + MbUb = MaVa + MaVa

Substituting values;

(0.08 kg * 0.5 m/s) + (0.05 kg * 0 m/s) = (0.08 kg * −0.1 m/s) + (0.05 kg * v)

0.4 = -0.008 + 0.05v

v = 8.16 m/s

Learn more about more about momentum: brainly.com/question/24030570:

#SPJ1

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Answer:

Good government

Explanation:

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3 years ago
El peso de María es de 617,4 N. Si en dos meses pierde 2,5 Kg de masa, ¿cuál será su nuevo peso?
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Two hundred and thirsty years
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Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
When the green arrow and solid red light is illuminated, __________?
otez555 [7]
<span>When the green arrow and solid red light is illuminated, </span>means you turn in the direction of the arrow.
5 0
3 years ago
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a
Levart [38]

Answer:

9.36*10^11 m

Explanation

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

5 0
2 years ago
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