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Sliva [168]
2 years ago
12

The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to

a pot. In the pot are red balls at the bottom and blue balls at the top. Black lines with arrowheads point upward from the red balls in the center of the pot, and black curved lines labeled W with arrowheads point downward from the blue balls along the sides of the pot. The handle of the pot is red near the pot and labeled X. The pot handle is held by a hand labeled Z. In which areas of the diagram does conduction occur? W and X X and Z Y and Z Z and W

Physics
2 answers:
kupik [55]2 years ago
8 0

X and Z is the best answer i could come up with.

Pepsi [2]2 years ago
6 0

Answer:

X and Z

Explanation:

Conduction occurs through direct physical contact.  Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.

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If you take any pitch on the keyboard, the next occurrence of the same letter name going towards the left (down) will vibrate:
NikAS [45]

Answer:

A. Twice as slow

Explanation:

8 0
2 years ago
En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2
Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F = 1961.4\,N

Txomin

F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F = 2157.54\,N

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0
2 years ago
A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

5 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
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Nataliya [291]

Answer:

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7 0
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