All categories

2 years ago

What is the force on a 52 coulomb charge in a field of 5 N/C

Physics

1 answer:

slega [8]2 years ago

8 0

Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf

Check out upload.

Download pdf

You might be interested in

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

Question number 8 Plz help

Dennis_Churaev [7]

To me, that sounds like the "Law of Conservation of Energy".

5 0

3 years ago

Oscillations of electricity and magnetism create ________________________ waves, which include visible light waves.

kaheart [24]

Explanation:

radio waves, which include visible light waves.

4 0

3 years ago

Mercury has a density of 13.56 g/mL. How many kilograms of mercury would you expect to fit in a cylindrical glass cup with a bot

vovangra [49]

Answer:

263.152kg

Explanation:

The density of a substance is related to its mass and volume as follows;

density = mass / volume

mass = density x volume -------------(i)

The substance in question here is mercury which has;

density = 13.56g/mL = 13.56g/cm³

Since the mercury is going to be put in the cylindrical glass, the volume of the cylindrical glass is going to be equal to the volume of the mercury that will be put.

And we know that the;

volume of a cylinder = πr²h

Where;

π = 3.142

r = bottom radius of the cylinder = 5.75inches

h = height of the cylinder = 0.950ft

For uniformity, let's convert the radius and height of the cylinder to their corresponding values in cm

r = 5.75 inches = 5.75 x 2.54 cm = 14.605cm

h = 0.950 ft = 0.950 x 30.48 cm = 28.956cm

Therefore, the volume of the cylinder;

v = 3.142 x (14.605cm)² x 28.956cm = 19406.5cm³

v = 19406.5cm³ [This is also the volume of the mercury necessary to fit the cylinder]

Now the following value has been found;

volume = 19406.5cm³

Substitute the values of density and volume into equation (i) as follows;

mass = 19406.5cm³ x 13.56g/cm³

mass = 263152.14g

Convert the result to kg by dividing by 1000

mass = 263.152kg

Therefore, 263.152kg kilograms of mercury would fit in the cylindrical glass.

3 0

3 years ago

How are metal bridges built to cope with changes of temperature?

nirvana33 [79]

As the metal expands as does the road bed so neither really effevts those foing over the bridge. as it is hot the metal will expand and so will most tarmac on roads.

8 0

2 years ago