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3 years ago

Hat is the approximate pressure of air at sea level?

Physics

2 answers:

Tanya [424]3 years ago

5 0

c) 101kPa

Hope I helped! ( Smiles )

Elenna [48]3 years ago

4 0

the answer is c.) 101kPa

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The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

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3 0

1 year ago

How do people participate in tourism

Makovka662 [10]

Community participation in tourism helps to uphold the local culture, tradition and indigenous knowledge of the local people. It also helps in conservation of the environment and culture of the local community.

5 0

3 years ago

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Degger [83]

Answer:

147.456077993 Hz

Explanation:

$f_0$ = Frequency of the sonar = 22 kHz

$v_w$ = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

$\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz$

The difference in frequency is 147.456077993 Hz

6 0

3 years ago

How would convection warm up an iguana cage

harina [27]

Using a basking spot so some sort of heated object, for example heating lamp or heating pad.

3 0

3 years ago

A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back

saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

$K.E =\dfrac{1}{2}mv^2$

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

$K.E=\dfrac{1}{2}\times0.5\times(4)^2$

$K.E = 4\ J$

The average force exerted by the floor on the ball = change in kinetic energy over collision time

$F = \dfrac{4}{0.05}$

$F=80\ N$

Hence, The force exerted by the floor is 80 N.

4 0

3 years ago