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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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In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which
Nesterboy [21]

Answer:

A

Explanation:

voltage between A and C is equal battery's voltage.

7 0
2 years ago
Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an
morpeh [17]

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = <em>Fvt cosθ</em>

<em></em>

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = <em>Fv cosθ</em>

8 0
1 year ago
The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?
n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100

The measured value is 2.02 meters and the actual value is 2.00 meters.

V_{measured}=2.02\\\\V_{true}=2.00

Percent Error=\frac{2.02-2.00}{2.00} *100

First, evaluate the fraction. Subtract 2.00 from 2.02

Percent Error=\frac{0.02}{2.00}*100

Next, divide 0.02 by 2.00

PercentError=0.01 *100

Finally, multiply 0.01 and 100.

Percent  Error=1\\Percent  Error= 1 \%

The percent error is 1%.

6 0
2 years ago
Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?
ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

  • When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0
2 years ago
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
ipn [44]

Answer:

ball clears the net

Explanation:

v_{o} = initial speed of launch of the ball = 20 ms^{-1}

\theta = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

v_{ox} = initial velocity = v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}

t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}       Eq-1

Consider the motion of the ball along the vertical direction

v_{oy} = initial velocity = v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}

t = time of travel

Y_{o} = Initial position of the ball at the time of launch = 2 m

Y = Final position of the ball at time "t"

a_{y} = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\

Since Y > 1 m

hence the ball clears the net

7 0
2 years ago
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