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3 years ago

6

A teacher stands well back from an outside doorway 0.88 m wide, and blows a whistle of frequency 780 Hz. Ignoring reflections, e

stimate at what angle(s) it is not possible to hear the whistle clearly on the playground outside the doorway.

1 answer:

AlladinOne [14]3 years ago

5 0

the wavelength is to be 56.67 cm and I know that i should be applying the double slit equation for destructive interference.

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The movement of water is able to transport minerals and nutrients. Which statement best explains why water is able to do this?

Marrrta [24]

no it can't do this why because I think that it is water and it can not go any where.

3 0

3 years ago

Read 2 more answers

Please indicate how long each bar is in centimeters and millimeters.

Maru [420]

Answer:

1) 74 cm or 740 mm

2) 33 cm or 330 mm

3) 76 cm or 760 mm

4) 17 cm or 170 mm

brainliest !?!?

6 0

2 years ago

A sample of helium behaves as an ideal gas as energy is

alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat $c_h = 5.193 J/g K$

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

$E_h = m_hc_h \Delta T = 20 J$

where $m_h$ is the mass of helium, which we are looking for:

$m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g$

4 0

3 years ago

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h

Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft$

The minimum stopping distance is 141.78 ft.

4 0

3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s$

In free fall

$v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m$

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

$s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s$

The time the rocket is accelerating is 6.75 seconds

5 0

3 years ago