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SSSSS [86.1K]
2 years ago
7

What is the most likely reason for some antelope to employ selective brain cooling

Physics
1 answer:
Anika [276]2 years ago
5 0

Answer:

The brain is a part of the body that is particularly sensitive to high temperature. Hence some ungulates, like the Thomson's gazelle, use a counter-current heat exchanging structure known as the carotid rete to keep the brain cooler than the body.The cooled arterial blood then continues toward the brain.

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A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The
kifflom [539]

Answer:

option C

Explanation:

The correct answer is option C                        

There is no external force acting in the system hence the momentum will be conserved.                                    

As the milk is leaking out of the tank mass of the tanker is decreasing.

When the mass of the container will decrease to  conservation the momentum speed of the container will have to be increased.

So, the car carrying milk will speed up.

8 0
3 years ago
Hey guys can you help me solve this problem "how long will it take a car travelling 30m/s to come to stop ifs its acceleration i
salantis [7]

Answer:

10 seconds.

Explanation:

We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time <em>t: </em>

<em />\displaystyle v_f = v_i + at<em />

<em />

The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².

Substitute and solve for <em>t: </em>

<em />\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\  t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}<em />

<em />

Hence, it will take the car 10 seconds to come to a stop.

7 0
2 years ago
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
4 years ago
A body starts from rest and travels ‘s’ m in 2nd second, than acceleration is.
Lana71 [14]

Answer:

3m

Explanation:

3 0
2 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
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