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3 years ago

The same 100-cm 3 lead block is carefully submerged in a container of mercury. One cm 3 of

Physics

1 answer:

omeli [17]3 years ago

8 0

Answer:

$100cm^3$

Explanation:

Using the $Displacement \ Method$ :-

If a solid mass of a known volume is submerged in a container of a fluid, the volume of the fluid displaced is equal to the volume of the solid mass.

Hence volume of mercury displaced is $100cm^3$

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A student investigated how the mass of water in an electric kettle affected the time taken for the water to reach boiling point.

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2 years ago

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2 years ago

a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes

kiruha [24]

The angle through which the grinding wheel rotates in the first second = <u>5300 rad</u>

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second = <u>5300 rad</u>

Learn more about angular velocity here:-brainly.com/question/6860269

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1 year ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers