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1 year ago

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A body of mass 250g is release from the top of a building. if the body hits the ground with a velocity of 4m/s, calculate th hei

ght of the building [g=10m/s²]

1 answer:

Nutka1998 [239]1 year ago

5 0

$\purple{\maltese}\large\underline{\underline{\sf\:\: Given :}} \\$

» Mass ( m ) of the body = 250 g

» final velocity of the body = 4 m/s

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: To \: \: Find :}} \\$

» The height of the Building= ??

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: Solution :}} \\ \\$

★ <u>Height of the </u><u>building </u><u>by the principal of conservatio</u>n;

$\\\\\begin{gathered}\purple{\maltese}\large\underline{\underline{\sf\:\: using \: formula :}} \\ \\\end{gathered}$

$\bigstar \: \underline{ \boxed{\sf { \pink{ \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: }}}}$

$\\ \: \: \underline{\textsf {Putting values \: in \: the \: \: given \: formula :}} \\$

$\\ \sf \implies \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: \\$

$\\ \sf \implies \: \frac{v {}^{2} }{2} \: \: = \: gh \: \\$

$\\ \sf \implies \: h \: = \frac{v {}^{2} }{2g} \: \: \: \\$

$\\ \sf \implies \: h \: = \frac{16 }{20} \: \: \: \\$

$\\ \sf \implies \: h \: = 0.8 \: \: \: m \\$

$\\ \sf \implies \: h \: = 80 \: \: \: cm \\$

henceforth , The height of the Building is <u>0.8 m or 80 cm</u> ..!!!

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What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

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3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

<em></em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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3 years ago

The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic

Gekata [30.6K]

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is = $\frac{0+40}{2\\}$ {initial velocity + final velocity}

= $\frac{40}{2}$

=20

Therefore, the magnitude of the average velocity of the car is 20 m/s

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3 years ago

What is a tool that can be used to measure the size of a force

Travka [436]

The answer to this question would be: a spring scale.

The spring scale that you use to determine your body weight is actually a device that measures your body gravitational force. The force itself influenced by your body weight, that is why it can determine your body weight.
More weight means more force, more force will shrink the spring more.

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3 years ago

Which of the following is the best thermal insulator?

Vera_Pavlovna [14]

B. Aluminum is possibly correct

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3 years ago