Visible light with a wavelength of 480 nm appears Question 33 options:

Can someone help me convert these?

fomenos

Assuming you are supposed to write each conversion in scientific notation:

(2) 1 m = 100 cm, so

(67 cm) × (1/100 m/cm) = 67/100 m = 0.67 m = 6.7 × 10 ⁻¹ m

(3) 1 km = 1,000 m, so

(1.2 km) × (1000 m/km) = 1200 m = 1.2 × 10³ m

(4) 1 m = 1,000 mm = 10³ mm, so

(6.2 × 10 ⁻³ m) × (10³ mm/m) = 6.2 mm

(5) 1 m = 1,000,000,000 nm = 10⁹ nm, so

(4.05 × 10³ nm) × (1/10⁹ m/nm) = 4.05 × 10 ⁻⁶ m

(6) 1 g = 1,000,000 µg = 10⁶ µg, so

(3200 µg) × (1/10⁶ g/µg) = 3200 × 10 ⁻⁶ g = 3.2 × 10 ⁻³ g

4 0

3 years ago

Acceleration is zero if

tester [92]

It’s doesn’t change meaning it’s 0

4 0

3 years ago

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the pro

Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) × (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

6 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago

State Newton's second law of<br>motion and write down<br>three examples to<br>justity it.

kakasveta [241]

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

$f = m \times a$

f= force

m=mass

a=acceleration

Explanation:

examples:

riding your bicycle

•your bicycle is the mass, your leg pushing in pedals of your bicycle is the force

pushing a box

•the box is the mass, you are pushing the box

setting a pencil down in a table

•the pencil is the mass, you are puting the pencil down

3 0

3 years ago