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2 years ago

-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck

Physics

1 answer:

Hoochie [10]2 years ago

6 0

Answer:

I would shout fore help if I was being raped or try to make him or her stop

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What is the path of an electron moving at 4 000 m/s perpendicular to a magnetic field of 1.5 T? (me = 9.11 x 10^-31kg)

morpeh [17]

Answer:

$15.18\times 10^{-9} m$

Explanation:

Radius of the path of electron which is moving perpendicular to magnetic field is defined as,

$r=\frac{mv}{qB}$

Here, m is the mass of electron, v is the velocity of electron, q is the charge on electron, and B is the magnetic field.

Given that, the velocity of electron is, $v=4000m/s$ .

And the magnetic field is $B=1.5T$ .

And the mass of electron is, $m=9.11\times10^{-31}kg$

And the charge on electron is, $q=1.6\times 10^{-19}C$

Put all these values in radius equation,

$r=\frac{9.11\times10^{-31}kg\times 4000m/s }{1.6\times 10^{-19}C\times 1.5T}\\r=15.18\times 10^{-9} m$

Therefore, the path radius of a moving electron is $15.18\times 10^{-9} m$

7 0

3 years ago

A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string

kupik [55]

Answer:

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1.6 x 0 + 0.8 x 0 = 1.6 x v₁ + 0.8 x 3

0 = 1.6v₁ + 2.4

-1.6v₁ = 2.4

v₁ = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

6 0

2 years ago

a block initially at rest has a mass m and sits on a plane incline at angle. it slides a distance d before hitting a spring and

KiRa [710]

Answer:

$k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}$

Explanation:

Let assume that spring reaches its maximum compression at a height of zero. The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}=U_{k,B} + W_{f}$

$m\cdot g \cdot (d + x_{f})\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x_{f}^{2}+\mu_{k}\cdot m \cdot g \cdot (d+x_{f})\cdot \cos \theta$

$m\cdot g \cdot (d + x_{f})\cdot (\sin \theta-\mu_{k}\cdot \cos \theta) = \frac{1}{2}\cdot k \cdot x_{f}^{2}$

The spring constant is cleared in the expression described above:

$k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}$

6 0

2 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$

6 0

3 years ago

Which optical device can focus light to a point through reflection?

frosja888 [35]

Answer:

Explanation:

did the quiz

7 0

2 years ago