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Vladimir79 [104]

3 years ago

6

When you walk with a cup of coffee (diameter5 cm) at just the

1 answer:

vladimir1956 [14]3 years ago

4 0

Answer:

10 cm/sec

Explanation:

using formula velocity = frequency * wave length

where: f=1/T

1/1=1

so frequency will be 1HZ ( due to 1 oscillation in 1 sec)

wave length will be 10 cm because while sloshing cup hold (1/2) half full length.

velocity = 1*10 = 10cm/sec.

velocity = 10 cm/sec

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What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

Paha777 [63]

$E = mgh + \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh + \frac{1}2} m r ^{2} \omega ^{2} + \frac{1}{2} I \omega^{2}$

for a solid cylinder: $I = \frac{1}{2} m r^{2}$
for a hollow cylinder: $I = mr^{2}$

I will look at the case of a hollow cylinder:

$E = mgh + I \omega ^{2} = constant \\ \\ I = \frac{mgh}{ \omega^{2} }$

That is as far as i get.

7 0

4 years ago

Particle-X has a speed of 0.720 c and a momentum of 4.350x1019 kgm/s. What is the mass of the particle? 2.0206 10-27 kg Hints: T

kirill115 [55]

Explanation:

Given that,

Speed of particle = 0.720 c

Momentum = 4.350\times10^{-19}\ kgm/s[/tex]

(I). We need to calculate the mass of the particle

Using formula of momentum

$P=mv$

$m =\dfrac{P}{v}$

$m=\dfrac{4.350\times10^{-19}}{ 0.720\times3\times10^{8}}$

$m=2.013\times10^{-27}\ Kg$

We need to calculate the rest mass of particle

Using formula of rest mass

$m=\dfrac{m_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}$

Where, $m_{0}$ = rest mass

Put the value into the formula

$m_{0}=2.013\times10^{-27}\times\sqrt{1-(\dfrac{0.720 c}{c})^2}$

$m_{0}=2.013\times10^{-27}\times\sqrt{1-(0.720)^2}$

$m_{0}=1.4\times10^{-27}\ kg$

(b). We need to calculate the rest energy of the particle

Using formula of energy

$E_{0}=m_{0}c^2$

Put the value into the formula

$E_{0}=1.4\times10^{-27}\times(3\times10^{8})^2$

$E_{0}=1.26\times10^{-10}\ J$

(c). We need to calculate the kinetic energy of the particle

Using formula of kinetic energy

$K.E=mc^2-m_{0}c^2$

$K.E=(m-m_{0})\timesc^2$

$K.E=(2.013\times10^{-27}-1.4\times10^{-27})\times3\times10^{8}$

$K.E=1.84\times10^{-19}\ J$

(d). We need to calculate the total energy of the particle

Using formula of energy

$E=mc^2$

Put the value into the formula

$E=2.013\times10^{-27}\times(3\times10^{8})^2$

$E=1.812\times10^{-10}\ J$

Hence, This is the required solution.

8 0

3 years ago

A man loads 260.5 kg of black dirt into his pickup. 60,452 g blows out on the ride home. How much black dirt does the man have w

Ivan

Answer:

200.048 kg

Explanation:

Total Black dirt loaded in the pickup (A) = 260.5 kg = 260.500 kg

Quantity of the dirt that blew away (B )= 60452 g = 60.452 kg

Remaining quantity of the black dirt is

$= 260.500 - 60.452$

$= 200.048$

Thus, the amount of black dirt the man has when he reaches home = 200.048 kg.

7 0

3 years ago

show that the centre of a rod of mass M and length L lies midway between the ends assuming that the rod has uniform cross sectio

Ivan

Explanation:

The cross section per unit length is uniform, so ρ is constant.

The center of mass is therefore:

x_avg = (∫₀ᴸ x ρ dL) / (∫₀ᴸ ρ dL)

x_avg = (∫₀ᴸ x dL) / (∫₀ᴸ dL)

x_avg = (½ L²) / (L)

x_avg = ½ L

8 0

3 years ago

I AM SO SORRY THIS TIME I PUT IT LOL

Fittoniya [83]

Answer:

a Internal core

b The lower mantle

c Butter dough

d Earth rocks

7 0

2 years ago