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2 years ago

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A sliver cylindrical rod has a length of 0.5 m and a radius of 0.4 m, find the density of the rod if it's mass is 2.640 kg

1 answer:

Serga [27]2 years ago

7 0

Answer:

Density=10.50kg/m³

Explanation:

$Solution,\\Height(h)=0.5m\\Radius(r)=0.4m\\Now, \\Volume=\pi r^{2} h\\\\Volume=\pi *(0.4m)^{2} *0.5m\\\\Volume=0.251327m^{3} \\\\Again,\\\\Density=\frac{mass}{volume} \\\\Density=\frac{2.640kg}{0.251327} \\\\Density=10.50kg/m^{3}$

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A bus took 8 hours to travel 639 km. For the first 5 hours, it

sergey [27]

Answer:

93

Explanation:

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3 years ago

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Anvisha [2.4K]

The answer is A)

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4 years ago

Read 2 more answers

Electric current in electric bulb is 20 mA, what is the charge flowing through the filament in 4 sec

anzhelika [568]

Answer:

The charge flowing through the filament is 0.08C

Explanation:

$I=20mA = 0.02A\\t= 4s\\$

Plug your given values into the charge and current formula:

$I=\frac{Q}{t} \\Q=I*t\\Q= (0.02A)(4s)\\Q= 0.08C$

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

¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *

lys-0071 [83]

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

Fuerza [force] (F) = 110 N
Metros [meters] (m) = 20 m
Trabajo [work] (W) = ?

Usar la fórmula (use formula):

$\boxed{\bold{W = F*d}}$

Reemplazar (replace):

$\boxed{\bold{W = 110\ N*20\ m}}$

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

$\boxed{\boxed{\bold{W =2.200\ J}}}$

Greetings.

7 0

3 years ago