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DaniilM [7]
2 years ago
15

A group of animals have short, narrow legs that are partly webbed. In which environment do these animals MOST LIKELY live? A) de

serts B) grasslands C) mountains D) rainforests
Physics
2 answers:
Burka [1]2 years ago
7 0
I'd say it's answer D) rainforests
Galina-37 [17]2 years ago
4 0

Answer:

D) rain-forests

Explanation:

Such animals live in rain-forests as they need to travel on land as well as in shallow waters. The geography of rain forests is such that it receives a lot of rain throughout the year. This makes it necessary for the animals to have such features in their legs to help them survive in that environment.

For other environment they would not need partly webbed legs.

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What is a <br> quadratic graph
Paladinen [302]

Answer:

The graph of a quadratic function is a parabola whose axis of symmetry is parallel to the y -axis. The coefficients a,b, and c in the equation y=ax2+bx+c y = a x 2 + b x + c control various facets of what the parabola looks like when graphed.

Explanation:

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3 years ago
Find the mean of set values 12g 9g 13g 12g 20g 17g 15g
nalin [4]

{\huge{\boxed{\mathcal{\green{Answer}}}}} \\ \frac{12 + 9 + 13 + 12 + 20 + 17 + 15}{7}  \\  =  \frac{98}{7}  \\  = 14 \\ {\huge{\boxed{\mathcal{\green{Hope \: it \: Helps}}}}}

5 0
2 years ago
a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
GalinKa [24]

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

∴acceleration, a= 14/70 = 0.2 m/s²

now on applying conservation of linear momentum

pi=pf            pi=0 (initially at rest)

0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s

v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

0- (0.3) ²= -2 x 0.2 x 5

s= 0.09/0.4 ≈ 0.3 m

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brainly.com/question/15172156

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6 0
1 year ago
Is O2 classified as a compound?
dybincka [34]

Answer: NO.

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3 0
3 years ago
Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,
Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules

Power

P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W

Converting to hp

1\ W=\frac{1}{745.7}\ hp

\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0
3 years ago
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