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liubo4ka [24]
2 years ago
15

Use the drop-down menus to complete the statements.

Physics
1 answer:
jekas [21]2 years ago
4 0

Answer:

Use the drop-down menus to complete the statements.

 

When electrons are lost, a  

✔ positive

 ion is formed.

When electrons are gained, a  

✔ negative

 ion is formed.

Explanation:

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While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density
bezimeni [28]
The equation for percent error is

% Error = 100*|Experimental-Theoretical|/Theoretical

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%
3 0
3 years ago
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h
Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion \alpha= 10.5\times10^{-6}\ K^{-1}

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

\Delta L=L\alpha\Delta T

Where, \Delta T= temperature difference

\alpha=Coefficient of thermal expansion

L= length

Put the value into the formula

\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})

\Delta L=0.42\ m

Hence, The difference in the length of the bridge is 0.42 m.

5 0
3 years ago
Who was the first to hypothesize that electron orbit a positively charged nucleus
Eva8 [605]
I think the answer is ruthorford
6 0
3 years ago
In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from
Sliva [168]

Answer:

spacing between the slits is 405.32043 ×10^{-9}  m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4I_{0} × cos²∅/2

so

I/4I_{0}  = cos²∅/2

here I/4I_{0} is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 ×cos^{-1} 0.921954

∅  = 45.56°

∅  = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅  = 2π d sinθ / wavelength

so

d = ∅× wavelength /  ( 2π  sinθ )

put all value

d = 0.795 × 610×10^{-9} / ( 2π  sin2.95 )

d = 405.32043 ×10^{-9}  m

spacing between the slits is 405.32043 ×10^{-9}  m

7 0
2 years ago
Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

8 0
3 years ago
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