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dlinn [17]
3 years ago
9

5. What happens to the arrangement of water molecules as ice melts?

Physics
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer: I am pretty sure the answer is B

Explanation: If not sorry bro.

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For the following questions consider a piece of copper wire. a. What type of bond is formed between the copper atoms? b. Describ
yarga [219]

Answer:

a. metallic bond

b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

c. due to the presence of free electrons in its outer energy levels

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3 years ago
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When an athlete holds a barbell overhead, the reaction force is the weight of the barbell on his hand. how does this force vary
Sonja [21]
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
3 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
PLEASE HELP I WILL GIVE MAX POINTS
Finger [1]

you can collect water and shine a light though it and record your finings

4 0
3 years ago
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A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast
Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15*10^{4}N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15*10^{4} N

3 0
3 years ago
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