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2 years ago

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A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no apprecia

ble air drag. How tall, in meters, is the building?

1 answer:

Elenna [48]2 years ago

6 0

Answer:A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no ..

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Calculate the potential energy of a rock that has a 45 kg mass and is sitting

Montano1993 [528]

Answer:

P.E = 13230 J

Explanation:

Given,

The mass of the rock, m = 45 Kg

The rock is sitting at a height from the ground, h = 30 m

The acceleration due to gravity, g = 9.8 m/s²

The potential energy of the body is given by the formula,

P.E = mgh joules

Substituting the given values in the above equation

P.E = 45 x 9.8 x 30

= 13230 J

Hence, the potential energy of the rock is P.E = 13230 J

6 0

3 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

There was an airplane crash, every single person on board died, but yet two people survived. How is this possible?

serg [7]

If everyone died then the too people were not on the plane they were near by when the plane crashed and were not onbard.

8 0

3 years ago

Read 2 more answers

Please answer fast!!!

sesenic [268]

Answer:1. A chair leaning on a wall

Explanation:

5 0

3 years ago

Read 2 more answers

A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in

Talja [164]

Answer:

132 N

Explanation:

Given that a 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail

From Newton 2nd law of motion,

Change in momentum = impulse.

Change in momentum = m( V - U )

Substitute all the parameters into the formula

Change in momentum = 1.1 ( 4.5 - 1.5 )

Change in momentum = 1.1 × 3

Change in momentum = 3.3 kgm/s

Impulse = Ft

That is,

Ft = 3.3

Substitute time t into the formula above

F × 0.025 = 3.3

F = 3.3 / 0.025

F = 132 N

Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.

3 0

3 years ago