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butalik [34]
3 years ago
10

Leta Stetter Hollingworth conducted pioneering work on __________. A. identity development in ethnic minorities B. cognitive pro

cesses in animals and humans C. racism and its effects on child development D. adolescent development and gifted children  
Physics
2 answers:
dalvyx [7]3 years ago
7 0

This is the only answer in the choices.

D. adolescent development and gifted children  

rusak2 [61]3 years ago
5 0
I believe the correct answer from the choices listed above is option D. Leta Stetter Hollingworth conducted pioneering work on adolescent development and gifted children. She <span>was an American psychologist who conducted pioneering work in the early 20th century. </span>
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Two samples of aluminum foil, 27 grams and 54 grams, are cut out from a roll. Which of the following properties is different for
Delicious77 [7]
Before answering this question, you must know the concept between extensive and intensive property. The extensive property does not depend on the amount of substance (like mass), which is the opposite of intensive properties. From the given choices, the rest are extensive properties except for <em>amount of matter</em>. Hence, that is the answer.
4 0
4 years ago
Read 2 more answers
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)
Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

P_{1}=\rho g h_{1}

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

P_{1}=\rho gh_{1}.....(I)

Pressure for second pipe,

P_{2}=\rho gh_{2}.....(II)

From equation (I) and (II)

P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

Put the value of P₁ and P₂

\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)

2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2....(III)

We know that,

The continuity equation

v_{1}A_{1}=v_{2}A_{2}

v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})

Put the value of v₂ in equation (III)

2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2

2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2

Here, \dfrac{A_{1}}{A_{2}}=\gamma

So, 2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)

v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

Hence, (a). The gauge pressure at the bottom of tube 1 is P_{1}=\rho g h_{1}

(b).  The speed of the fluid in the left end of the main pipe \sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}

8 0
3 years ago
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo
BigorU [14]

Answer:

\lambda = 6.25\times10^{-9}= 625 nm

Explanation:

We now that for

for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....

d= distance between the slits, λ= wavelength of incident ray

for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.

Given

d=1.19 mm, y=4.97 cm,  and,   n=10,   D=9.47 m

applying formula

λ= (d*y)/(D*n)

putting values we get

\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}

on solving we get

\lambda = 6.25\times10^{-9}= 625 nm

8 0
3 years ago
Which determines the pitch of a sound? A. At the same wavelength, the sound with the most cycles per second will have the highes
kipiarov [429]
This would be known to be determined by its sound wave which would be the amplitude. The pitch of sound would actually be a little differenr in this matter. But know that we know and understand the sound wave of the pitch of sound, we know that this is done by amplitude.
5 0
4 years ago
An astronaut is moving in space when a big explosion occurs about 50 meters behind him. How will the astronaut come to know abou
LUCKY_DIMON [66]

Answer:

The correct answer is B.

The astronaut will know due to the light from the explosion.

Explanation:

Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.  

Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.

Cheers!

7 0
3 years ago
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