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3 years ago

Which process has led to the production of most modern domesticated crops and livestock

Physics

2 answers:

Nesterboy [21]3 years ago

8 0

The Answer is A: Selective Breeding.

boyakko [2]3 years ago

6 0

A selective breeding. Put your best animals together to get better offspring.

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What kind of friction occurs as a fish swims through water?

Paul [167]

It must be sliding friction, because the fish is already in motion.

7 0

3 years ago

True or false, the older rocks tend to be closer to the tear in the ocean floor

Natali5045456 [20]

Answer:

false!!

your welcome!!!!

4 0

2 years ago

A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.

Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0

2 years ago

What is an expression for the vertical component of this vector?

grandymaker [24]

The correct choice is D .

4 0

2 years ago

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its

s344n2d4d5 [400]

Answer:

Part a)

Moment of inertia of the cylinder is given as

$I = 1.21 \times 10^{-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

Explanation:

As we know that the inclined plane is of length L = 3 m

and its inclination is given as 25 degree

so we know that acceleration of center of mass of the cylinder is constant so we will have

$v_f^2 = v_i^2 + 2 a L$

so we have

$v_f^2 = 0 + 2a(3)$

now we know that

$v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}$

$\frac{3}{1.50} = \frac{v_f + 0}{2}$

$v_f = 4 m/s$

Now we have know that final speed of the cylinder due to pure rolling is given as

$v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}$

$4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}$

$I = 1.21 \times 10^[-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

8 0

2 years ago