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Anna11 [10]
3 years ago
10

Stars that are composed of matter in which electrons have combined with protons are called ________.

Physics
1 answer:
Dovator [93]3 years ago
3 0
The answer will be neutral because the same number of protons and the same number of electrons

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Select the correct statement about osmolarity. a) Osmolarity measures the moles of solute per liter of solution. b) The contents
8090 [49]

Answer:

a) Osmolarity measures the moles of solute per liter of solution.

Explanation:

Osmolarity is defined as the number of moles of solute that contribute to the osmotic pressure, per liter of solution, of solution. That is, the measurement of the solute concentration. The prefix "osmo-" indicates the possible variation of the osmotic pressure in the cells, which will occur when the solution is introduced into the body.

7 0
3 years ago
An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

a). The work done by the tension is:

W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
2. Draw a ray diagram that shows the image
Lunna [17]

you have to draw diagram

6 0
3 years ago
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
What do electrons move from
Alona [7]

Answer:

Negatively charged, to positively charged parts

Explanation:

Electrons are negative, negative is attracted to positive.

7 0
3 years ago
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