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Karo-lina-s [1.5K]

3 years ago

6

Do deaf children go through the four stages of acquiring language? Use what you’ve read in the chapter to explain why or why not

. Further, explain any differences and/or similarities that may take place.

1 answer:

sdas [7]3 years ago

4 0

No deaf children dont. im deaf and i dident go through those stages.

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In which of these examples is the greatest movement occurring?

Elenna [48]

Answer:

not clear pic...but it's definitely not A)

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3 years ago

What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of

Arlecino [84]

It has to be Earthquake

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3 years ago

Read 2 more answers

Identify a true statement about flow separation. Multiple Choice A. Flow separation is caused due to high temperature gradient i

Helga [31]

Answer:

D. Flow separation is caused due to adverse pressure gradient in the flowing fluid.

Explanation:

Flow separation :

When adverse pressure gradient become dominate then phenomenon of flow separation occurs.In the other words when boundary layer is form against the adverse pressure then phenomenon of flow separation occurs.The adverse pressure means a opposing which act in the opposite to the direction of fluid flow.Due to flow separation eddy formation occurs and these eddy leads to increases the losses in the fluid flow.Due to flow separation fluid leaves the solid surface and form eddies.

So the answer is D.

6 0

3 years ago

An object is traveling at a constant velocity of 12 m/s when it experiences a constant acceleration of 2.5 m/s 2for a time of 20

faltersainse [42]

Answer:

62m/s

Explanation:

If the object accelerates at a constant rate of 2.5 m/s^2 for 20 seconds, then it will speed up by a total of 2.5*20=50m/s. Adding this to the initial velocity of 12m/s, you get a total velocity of 62m/s. Hope this helps!

6 0

3 years ago

A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant

Margarita [4]

Answer:

a) $W=0.8J$

b) $d_t=0.20m$

c) $\triangle K.E=0.267J$

Explanation:

From the question we are told that:

Mass of cart 1 $M_1=1.0kg$

Mass of cart 1 $M_2=0.05kg$

Force on cart 1 $F_1=4.0N$

Push Distance of cart 1 $d_1=0.20m$

Slide Distance of cart 1 $d_1'=0.35m$

a)

Generally the equation for work-done is mathematically given by

$W=f*d\\W=4*0.20\\W=0.8J \\$

b)

The systems center of mass moved a net totally of (while being pushed)

Mass 1 =0.20m

Mass 2=0

Therefore

$d_t=d_1+d_2$

$d_t=0.20+0$

$d_t=0.20m$

c)

Since work-done is equal to K.E energy of cart 1

Therefore

$W=1/2mv^2$

$V_1=\sqrt{\frac{W}{1/2m}}$

$V_1=\sqrt{\frac{0.8}{1/2(1)}}$

$V_1=1.264$

Therefore Kinetic energy before collision is

$K.E_1=1/2mv^2$

$K.E_1=1/2*1*1.264^2$

$K.E_1=0.768$

Generally from the equation for conservation of momentum the Velocity of cart 2 is mathematically given by

$v_2=\frac{m_1V_1}{m_1+m_2}$

$v_2=\frac{1*1.264}{1+0.5}$

$V_2=0.842m/s$

Therefore the final K.E is mathematically given by

$K.E_2=(1/2)(m_1+m_2)V_2^2$

$K.E_2=1/2*(1.5)(0.842)^2$

$K.E_2=0.531J$

Generally the Change in K.E is mathematically given by

$\triangle K.E=K.E_1-K.E_2$

$\triangle K.E=0.798-0.531$

$\triangle K.E=0.267J$

Therefore the will force change the kinetic energy of the system's center of mass by

$\triangle K.E=0.267J$

4 0

3 years ago