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2 years ago

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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

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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A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago

Who was theJewish patriarch asked to sacrifice his son

juin [17]

<h2>Answer:</h2>

Abraham

<h2>Explanation:</h2>

This story can be found in the bible, in Genesis 22. According to the story, the Lord tested Abraham's faith asking him to sacrifice Isaac his son in a mountain in the land of Moriah. Abraham traveled with his son three days until he found the place the Lord has established for the sacrifice. There, Abraham tied his son and laid him on the altar on top of a wood he had prepared. Abraham picked up a knife to kill his son as a sacrifice but the Lord called him from heaven saying: <em>"Do not hurt him in any way, for now I know that you truly fear God. You have not withheld from me even your son, your only son"</em>

7 0

2 years ago

A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?

Andrei [34K]

Answer:

(a) 30 m/sec

(b) -50 m/sec

Explanation:

We have given initial velocity of ball u = 50 m/sec

Acceleration due to gravity $g=-10m/sec^2$

(a) Time t = 2 sec

Now according to first equation of v = u-gt

So v=50-10×2=30 m/sec

(b) Time t = 10 sec

Now according to first equation of motion

So final velocity v = u-gt = 50-10×10 =-50 m/sec

3 0

3 years ago

Which subgenre would include stories about talking animals or magical powers? Image result for fable animal O fable O realistic

Furkat [3]

Answer:

fable

Explanation:

8 0

2 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

2 years ago