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2 years ago

What unit is used to measure wavelength?

Physics

2 answers:

irina [24]2 years ago

5 0

The unit used to measure wavelength is a Nano-meter

Mashcka [7]2 years ago

3 0

Answer: Nanometer .

Explanation: Hope this helped :)

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An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
So, we can write the following expression for Fg:

$F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)$

The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

$F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N (2)$

According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

$a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2 (3)$

3 0

2 years ago

calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m

bulgar [2K]

Answer:

$\huge\boxed{\sf Work\ done = 4 kJ}$

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

$\rule[225]{225}{2}$

5 0

1 year ago

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of

Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0

2 years ago

In designing buildings to be erected in an area prone to earthquakes, what are the things that you need to consider?

il63 [147K]

Answer:

In an earthquake prone area, the things that are needed to be considered for the buildings to stand constant and no cause of damages are as follows-

The foundation of the building must be made strong, using large amount of concretes so that it can hold the entire load of the building.

The pillars of the buildings are also should be firmed, rigid, strong and thick.

The buildings are needed to be construct in such a way that during any seismic event, the forces can be redistributed easily through the body of the building without causing any damage.

The buildings must be overall comprised of holding high strength and stiffness.

6 0

2 years ago

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

iogann1982 [59]

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

4 0

3 years ago