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xeze [42]
3 years ago
10

Deserts (especially true deserts) are not easy places for animals to live. Animals who live in them often have special features

that
help them survive. Which adaptation would be helpful for this group of animals?
keen eye sight
short, narrow legs
tolerate heat
webbed feet
Physics
1 answer:
mart [117]3 years ago
5 0

Answer:

tolerate heat

Explanation:

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A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele
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  • initial velocity=u=0m/s
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\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}

\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}

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What is the formula to find PE and KE ?
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PE stands for Potential Energy. It is the stored energy in an object due to its position with respect to some reference. It is expressed in Joules.

P.E = m * g * h    OR  P.E. = mgh

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g - acceleration due to gravity
h - height attained due to the body's displacement.

K.E. stands for Kinetic Energy. It is the energy possessed by a body due to its motion.

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2 years ago
How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it
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Explanation:

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How is work calculated when the force applied is not parallel to the displacement
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7 0
3 years ago
Read 2 more answers
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
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