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3 years ago

5

Consider electrons accelerated using a potential difference of (12.5 A B) kV before hitting a metal surface. Calculate the minim

um wavelength of the emitted x-rays. Give your answer in pm and with 3 significant figures.

1 answer:

Maslowich3 years ago

4 0

Answer:

Explanation:

Calculate The Minimum Wavelength Of The Emitted X-rays. Give Your Answer In Pm And With 3 Significant Figures. This problem has been solved! See the

Hope this helps! Have great day!

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A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

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

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

3 years ago

While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba

Yuri [45]

Answer a would be correct since velocity is a vector and has a magnitude and a direction. In this case v₁ = - v₂.

8 0

3 years ago

Read 2 more answers

Cuales son las ventajas de la botella de leyden

xxTIMURxx [149]

Answer-

Cono

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7 0

3 years ago

in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

baherus [9]

The initial speed of the vehicle before the collision is 8 m/s.

Let the mass of the vehicle = m
Let the initial speed of the vehicle stopped = u
The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>

The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s$

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

6 0

2 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago