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3 years ago

Newton’s third law of motion says that for every action there is a(n) and opposite reaction.

Physics

1 answer:

sattari [20]3 years ago

3 0

Answer:

for every action thete is an equal and opposite reaction

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A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

$f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

$T=4f^2lm$

$T=4(120)^2\times 0.75\times 0.012$

T = 518.4 N

Tension in the wire, T = m' g

$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0

3 years ago

What is perfect machine?Why is this machine not possible in the real life?

andrezito [222]

Explanation:

The machine whose efficiency is 100% is known as perfect machine .This machine is not possible in real life because every machine is affected by the overcoming friction due to which is efficiency become less than hundred percent .

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3 years ago

Which formula is used to find an object's acceleration?

irga5000 [103]

Acceleration= change in velocity/time

8 0

3 years ago

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Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m

astra-53 [7]

The answer is d i think

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3 years ago

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago