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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

1 answer:

mart [117]3 years ago

5 0

Answer:

tolerate heat

Explanation:

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a hammer of mass 5 kg travelling at 4 metre per second is a nail directly and does not rebound what is the impulse of the hammer

dmitriy555 [2]

20 kg.m/s
p =m*v = 5*4 = 20

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An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work

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Answer:

Its answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

71 ÷ 32 = m

2.21 kg = m

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Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

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The answer is C, because they moved from a stand still to down the hill

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3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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3 years ago

a red ball moves horizontally in a 30 m long tube what is the displacement of the red ball between 0s and 24s?

gayaneshka [121]

Answer:

30 metres.

Explanation:

Given that a red ball moves horizontally in a 30 m long tube.

Displacement is the distance travelled in a specific direction. It has both magnitude and direction.

Since the motion is horizontal, it moves is a certain direction.

Within the stipulation of time, the displacement will be the distance covered in the horizontal direction which is 30 metres.

Therefore, the displacement of the motion of the red ball is 30 metres.

7 0

3 years ago