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3 years ago

A grapefruit falls from a tree and hits the ground 0.72 s later. How far did the grapefruit drop?

Physics

1 answer:

ycow [4]3 years ago

7 0

Answer: 2.541884 meters

Explain:

Formula: s = 0.5 * g * t^2

Time of fall (t) is 0.72 while the gravitational acceleration rate is (9.80665m/s^2).

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The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu

zhenek [66]

Answer:

$f=2887.93Hz$

Explanation:

Given data

Length L=2.9 cm=0.029m

Speed of sound v=335 m/s

to find

Fundamental frequency f

Solution

As we know that frequency is given as:

f=v/λ

$f=\frac{v}{4L}\\ f=\frac{335m/s}{4(0.029m)} \\ f=2887.93Hz$

3 0

3 years ago

A 15-N force and a 45-N force act on an object in opposite directions

lidiya [134]

Answer:

30-N

Explanation:

Because they are both acting on an object going opposite directions, then you simply subtract the 15 from 45 as 45N of force is stronger than 15N of force.

8 0

2 years ago

Supercells are updrafts that may cause which of the following?

ale4655 [162]

just for anyone looking for the answer i just took the test and the answer is tornadoes

6 0

3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus

jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

f = 468.4 Hz

6 0

3 years ago