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2 years ago

IDENTIFY WHAT IS BEING DESCRIBED IN EACH SENTENCE AND WRITE YOUR ANSWER ON THE BLANKS

1 answer:

3 0

1) digestive enzymes

2) skeletal system

3) The integumentary system

4) the cardiovascular system

5) secrete hormones

Hope this helps! Have a good day! ❤️

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How do you find the velocity of an object?

Scorpion4ik [409]

<span>Basically just divide the change in position by the change in time. </span>

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3 years ago

What is inelastic deformation?

EleoNora [17]

The answer is A. Deformation is where the object doesnt return to its original shape.

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3 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

$v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}$

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

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3 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

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3 years ago

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago