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3 years ago

Based on its location on the periodic table, which element could be used for

Physics

2 answers:

Triss [41]3 years ago

6 0

C. Aluminum (Al)

Hope I helped you

user100 [1]3 years ago

3 0

Answer: Aluminum (Al)

Explanation: A P E X

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Two technicians are discussing how a clutch disengages. Technician A says a gap on each side of the clutch disk facing when dise

victus00 [196]

Answer: A - a gap on each side of the clutch disk facing when disengaged

Explanation:

A clutch switch is used to ensure the clutch is disengaged

or Prevent the engine from starting unless the clutch pedal is depressed

When a clutch is disengageda gap will be on each side of the facing.

5 0

3 years ago

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Which of the following devices could be used to create a position--time graph for an object?

pickupchik [31]

D none of the above

all of them is related to energy and motion , not to positioning.

6 0

4 years ago

During normal conditions, we only see the Sun’s __________, but during a solar eclipse, we get to see the _______________.

liberstina [14]

I would say A. because during the solar eclipse you cannot see the suns Photosphere at all, and the corona is the light that emits around the moon during the eclipse.

8 0

3 years ago

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Megan walks 1100 m to the left in 330s. What was her average speed in m/s?

Hunter-Best [27]

Answer:

3.33m/s

Explanation:

Speed is the change in distance of a body with respect to time.

Speed = Distance/Time

Given

Distance = 1100m

Time = 330s

Speed = 1100/330

Speed = 3.33m/s

Hence her average speed in m/s is 3.33m/s

7 0

4 years ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

7 0

4 years ago