Geography: In the diagram shown to the right, m

In the diagram shown to the right, m, №14523102, 15.12.2022 20:41

Biology

Study the diagram shown. (diagram below) which type of evidence for the theory of evolution does the diagram show? a homologous structures b embryological structures c bestial structures d analogous structures

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P Answered by Master

82

those are homologous structures, they share a common ancestor, they have similar bones but different functions

Mathematics

A. look at the diagram of the two squares shown below. the pythagorean theorem says that for any right triangle, the square of the length of the hypotenuse, c, is equal to the sum of the squares of the lengths of the legs (a and b): a2 + b2 = c2. explain how the diagram shown above can be used to prove the pythagorean theorem. show your work. b. look at the small box shown below. what is the length of the diagonal of the box shown? leave your answer in radical form. show your work and explain your steps.

Step-by-step answer

P Answered by PhD

5

Squares 1 and 2 have the same area because they're both (a+b) on a side.

Each of these squares is covered by four identical right triangle tiles, legs a,b, hypotenuse c.

In the first picture we see the uncovered part of the square, not covered by triangular tiles, is two squares, area a^2+b^2 .

In the second picture the uncovered part of the square is a smaller square, area c^2 .

We just moved the tiles around on the square, so the uncovered part is the same in both cases. So

a^2 + b^2 = c^2

--------

The rectangular prism has a bottom rectangular base 6 by 8. So the diagonal is $\sqrt{6^2+8^2}=\sqrt{100}=10$ .

The diagonal and the 7 cm side make a right triangle whose hypotenuse is the diagonal of the rectangular prism we seek.

$\sqrt{10^2 + 7^2} = \sqrt{149}$

√149

Biology

Study the diagram shown. (diagram below) which type of evidence for the theory of evolution does the diagram show? a homologous structures b embryological structures c bestial structures d analogous structures

Step-by-step answer

P Answered by Specialist

82

those are homologous structures, they share a common ancestor, they have similar bones but different functions

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Mathematics

A. look at the diagram of the two squares shown below. the pythagorean theorem says that for any right triangle, the square of the length of the hypotenuse, c, is equal to the sum of the squares of the lengths of the legs (a and b): a2 + b2 = c2. explain how the diagram shown above can be used to prove the pythagorean theorem. show your work. b. look at the small box shown below. what is the length of the diagonal of the box shown? leave your answer in radical form. show your work and explain your steps.

Step-by-step answer

P Answered by PhD

5

Squares 1 and 2 have the same area because they're both (a+b) on a side.

Each of these squares is covered by four identical right triangle tiles, legs a,b, hypotenuse c.

In the first picture we see the uncovered part of the square, not covered by triangular tiles, is two squares, area a^2+b^2 .

In the second picture the uncovered part of the square is a smaller square, area c^2 .

We just moved the tiles around on the square, so the uncovered part is the same in both cases. So

a^2 + b^2 = c^2

--------

The rectangular prism has a bottom rectangular base 6 by 8. So the diagonal is $\sqrt{6^2+8^2}=\sqrt{100}=10$ .

The diagonal and the 7 cm side make a right triangle whose hypotenuse is the diagonal of the rectangular prism we seek.

$\sqrt{10^2 + 7^2} = \sqrt{149}$

√149

Biology

The diagram shown represents a cell that will undergo mitosis. Which diagrams below best illustrate the nuclei of the daughter cells that result from a normal mitotic cell division of the parent cell shown?

Step-by-step answer

P Answered by PhD

it looks like a mad face hehehehe

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Biology

The diagram shown represents a cell that will undergo mitosis. Which diagrams below best illustrate the nuclei of the daughter cells that result from a normal mitotic cell division of the parent cell shown?

Step-by-step answer

P Answered by PhD

it looks like a mad face hehehehe

Geography

brainliest and 15 points Micky made the following diagram shown to represent the lunar eclipse. Earth is shown as a circle with water and land represented on it. On its left is a smaller circle labeled as Moon. On the left of the moon is the sun. Rays are shown from sun towards the moon and Earth and the shadow of the moon falls on Earth. Why is Micky s diagram of the lunar eclipse incorrect? Earth should be between the sun and moon. The sun should be between the moon and Earth. Earth should be located vertically above the sun. The sun should be

Step-by-step answer

P Answered by Master

19

Earth should be between the sun and moon.

Explanation:

Geography

brainliest and 15 points Micky made the following diagram shown to represent the lunar eclipse. Earth is shown as a circle with water and land represented on it. On its left is a smaller circle labeled as Moon. On the left of the moon is the sun. Rays are shown from sun towards the moon and Earth and the shadow of the moon falls on Earth. Why is Micky s diagram of the lunar eclipse incorrect? Earth should be between the sun and moon. The sun should be between the moon and Earth. Earth should be located vertically above the sun. The sun should be

Step-by-step answer

P Answered by Specialist

19

Earth should be between the sun and moon.

Explanation:

In the diagram shown to the right, m

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