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3 years ago

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With what tension must a rope with length 2.20 m and mass 0.150 kg be stretched for transverse waves of frequency 43.0 hz to hav

e a wavelength of 0.800 m ?

1 answer:

Blababa [14]3 years ago

5 0

Solution:

$v=sqrt(\frac{t}{u})\\v=40\times0.80=32m/s\\\frac{t}{u}=900\\t=900\times\frac{0.150}{2.20}\\=61.36N$

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Answer:

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Explanation:

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2 years ago

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A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?

Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

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1 year ago

A book is sliding across a horizontal desk and comes to a stop. the books kinetic energy was converted into what type of energy

Oliga [24]

Friction stole the book's kinetic energy, and turned it into heat energy ... which blew away in the breeze.

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3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

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2 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

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3 years ago