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Brrunno
2 months ago
8

Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t

he cart is M, and that of student 2 and the cart is 1.5M. If student 1 pushes student 2 so that she recoils with velocity v,  what is the velocity of student 2.
Physics
1 answer:
inna [3.1K]2 months ago
6 0

Answer:

  v₂ = v/1.5 = 0.667 v

Explanation:

We will apply the conservation of momentum for this problem, defining a system that includes both students and their carts; the forces during contact are considered internal, ensuring momentum is conserved.

Initial momentum before the push

p₀ = 0

Final momentum after the push yields

    p_{f} = m₁ v₁ + m₂ v₂

   0 = m₁ v₁ + m₂ v₂ p_{f}

   m₁ v₁ = - m₂ v₂

Substituting yields

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

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An apparatus is used to prepare an atomic beam by heating a collection of atoms to a temperature T and allowing the beam to emer
inna [3103]

Answer:

As indicated in the attached document

Explanation:

The comprehensive steps, mathematical reasoning, and manipulations are presented in the attachment.

3 0
2 months ago
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
Sav [3153]

Answer:

The electric flux going through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε = 8.8542 * 10^{-12}

Substituting ε = 8.8542 * 10^{-12} and q =7.6\µC; the formula simplifies to

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Thus, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
4 months ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
serg [3582]

Respuesta:

P_(bomba) = 98,000 Pa

Explicación:

Se nos proporciona;

h2 = 30m

h1 = 20m

Densidad; ρ = 1000 kg/m³

Primero, entendemos que la suma de las presiones en el tanque y la bomba es igual a la del boquilla,

Así, se puede expresar como;

P_(tanque)+ P_(bomba) = P_(boquilla)

Ahora, la presión se daría como;

P = ρgh

Y así,

ρgh_1 + P_(bomba) = ρgh_2

<ppor lo="" tanto="">

P_(bomba) = ρg(h_2 - h_1)

<pal sustituir="" los="" valores="" pertinentes="" obtenemos="">

P_(bomba) = 1000•9.8(30 - 20)

P_(bomba) = 98,000 Pa

</pal></ppor>
5 0
3 months ago
A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay
Sav [3153]

Answer:

v = 5.72\ m/s

Explanation:

Provided data:

The motorist's initial speed is 20 m/s.

The distance to the traffic light is 200 m.

The duration of the red light is 15 s.

It takes 1 s for the driver to apply the brakes.

Distance covered during this response time

d = s x t

d = 20 x 1 = 20 m

The time to arrive at the red light after factoring in the reaction time is 15 - 1 = 14 s.

The remaining distance to cover within this time is 200 - 20 = 180 m.

To determine the motorist's speed upon reaching the red light, employ the equation of motion

s = u t + \dfrac{1}{2}at^2

180 =20\times 14 + \dfrac{1}{2}\times a\times 14^2

98 a = -100

a = -1.02\ m/s^2

Utilizing the equations of motion to ascertain the velocity:

v^2 = u^2 + 2 a s

v^2 = 20^2 - 2\times 1.02\times 180

v = 5.72\ m/s

The speed of the motorist as she reaches the red light is 5.72 m/s.

}]} 2023-10-14 16:00:00.000000
7 0
4 months ago
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