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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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An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use
shtirl [24]

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object  = 200 feet/second

Final velocity of object  = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

a=\frac{v_f-v_i}{t}

where vf represents final velocity, vi represents initial velocity and  is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a= -30m/s

The acceleration of the object is -30m/s  

3 0
3 years ago
Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
To multiply the input force of a hydraulic lift , the input end should be the one having the
nikdorinn [45]
The smaller surface area

Hope this helps!
5 0
3 years ago
Please help!!!!!! Motion and Forces​
aniked [119]
157.5J
KE=1/2 mv^2
KE= 1/2(35kg)(3m/s)^2
KE=(17.5kg)(9m^2/s^2)
KE= 157.5J
4 0
3 years ago
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