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Alika [10]
3 years ago
10

One of the world trade center towers WTC1 is 1,368 feet tall, what is it's height in meters?​

Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Hello!

I'm just going to ignore the name of the building and just do the math

1 foot is equivalent to 0.3048 meters

if the building is 1368 feet tall, multiply it by 0.3048 to get 416.9664 meters

I hope this helps, and have a nice day!

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Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D​
erma4kov [3.2K]

Answer:

The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.

Kinetic energy is the energy that a body possesses due to its motion.

Potential energy is the energy a body possesses due to its position.

From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.

In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.

Therefore, the car C has KE = 100, PE = 0

6 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is
madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book(m_0)=1.7 kg

height(h)=1 m

time taken=0.64 s

h=ut+frac{at^2}{2}

1=0+\frac{a(0.64)^2}{2}

a=4.88 m/s^2and [tex]a=\alpha r

where \alphais angular acceleration of pulley

4.88=\alpha \times 5.2\times 10^{-2}

\alpha =93.84 rad/s^2

And Tension in Rope

T=m(g-a)

T=1.7\times (9.8-4.88)

T=8.364 N

and Tension will provide Torque

T\times r=I\cdot \alpha

8.364\times 5.2\times 10^{-2}=I\times 93.84

I=0.463\times 10^{-2} kg-m^2

I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0
3 years ago
a snail takes 16 minutes 40 seconds to cover a distance of 1 m calculate the average speed of the swimmer​
matrenka [14]
The snail’s speed is 0.001042. Hope this helps!

8 0
2 years ago
Read 2 more answers
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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