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netineya [11]
4 years ago
13

I

Physics
2 answers:
Vlad1618 [11]4 years ago
8 0
C. thalamus
All sensory information coming into the brain from the body must first pass thru the thalamus
Sati [7]4 years ago
8 0

Answer:

The correct answer is cerebrum.

Explanation:

Just took a quiz on this.

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What is the potential energy of the bowling ball as it sits on top of a building ?​
tangare [24]

Answer:

The potential energy of the bowling ball will be mgh

Explanation:

Let the mass of bowling ball =m

The height of building on which bowling ball sits=h

So,

The potential energy of the bowling ball =P.E.= mgh

8 0
3 years ago
When a cold air mass catches up with a warm air mass, the resut is often<br> a(n)<br> front.
Alex777 [14]

Answer:

a cold front i think

Explanation:

4 0
3 years ago
A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge
Kamila [148]

Answer 1) The electric field at distance r from the thread is radial and has magnitude  

E = λ / (2 π ε° r)  

The electric field from the point charge usually is observed to follow coulomb's law:  

E = Q / (4  π ε° r^{2})  

Now, adding the two field vectors:  

E_{thread}  =  {2.5 / (22 π ε° X 0.07 ) ; 0}  

Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

5 0
3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

6 0
3 years ago
the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change
defon
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=mC_s \Delta T
where
m is the mass of the substance
C_s the specific heat capacity
\Delta T the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is C_s = 4.2 J/g ^{\circ}C and the amount of heat supplied is Q=31500 J, so if we re-arrange the previous formula we find the increase in temperature of the water:
\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C
7 0
3 years ago
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