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3 years ago

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A 2.6 mm -diameter sphere is charged to -4.5 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm

of the surface of the target before being reflected. Part A What was the electron's initial speed

1 answer:

motikmotik3 years ago

7 0

Answer: $1.96\times 10^{8}\ m/s$

Explanation:

Given

Diameter of sphere is $d=2.6\ mm\quad \quad[r=1.3 mm]$

Charge on the sphere is $Q=-4.5\ nC$

Nearest distance electron can reach to sphere is $d=0.37\ mm$

Here, kinetic energy of electron is converted into electrostatic energy between the two i.e.

$\Rightarrow \dfrac{1}{2}mv^2=\dfrac{kQq}{d}\\\\\Rightarrow \dfrac{1}{2}\times 9.1\times 10^{-31}\times v^2=\dfrac{9\times 10^9\times 4.5\times 10^{-9}\times 1.6\times 10^{-19}}{(3.7\times 10^{-4})}\\\\\Rightarrow v^2=3.8491\times 10^{16}\\\Rightarrow v=1.96\times 10^{8}\ m/s$

Thus, the initial speed of electron is $1.96\times 10^{8}\ m/s$ .

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While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at

Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0

3 years ago

A 1400 kg aircraft going 40 m/s collides with a 1500 kg aircraft that is parked and they stick together after the collision and

Leni [432]

Answer:

67.6 m

Explanation:

For this problem, we just need to analyze the motion of the two planes after the collision.

The two planes stick together and move with a velocity of

u = 19.3 m/s

The problem tells us that they skid until they stop: so, their final velocity is

v = 0

Since the velocity is not constant, this means that they are accelerating.

We also know that they skid for a time of

t = 7 s

Therefore, we can find their stopping distance using the following suvat equation

$s=(\frac{u+v}{2})t$

And substituting, we find

$s=(\frac{0+19.3}{2})(7)=67.6 m$

6 0

3 years ago

in terms of mechanical advantage and velocity ratio write an expression for the efficiency of a simple machine

mixer [17]

Answer:

Efficiency = (MA/VR) ×100%

8 0

3 years ago

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago