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1 year ago

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

Physics

1 answer:

Debora [2.8K]1 year ago

7 0

You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.

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The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b

Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0

1 year ago

Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

Alex73 [517]

Explanation:

Below is an attachment containing the solution.

4 0

1 year ago

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0

2 years ago

If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal

iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0

1 year ago

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro

netineya [11]

Refer to the diagram shown below.

v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ = the angle that the string makes with the vertical
m = Bob's mass (mg = the weight)
F = centripetal force
l = the length of the string

From geometry,
r = l sin θ

The centripetal acceleration is
$a= \frac{v^{2}}{r} = \frac{v^{2}}{l \,sin \theta}$
The centripetal force is
$F = \frac{m v^{2}}{l \, sin \theta}$

For vertical force balance,
T cosθ = mg (1)
For horizontal force balance,
$Tsin \theta = F = \frac{m v^{2}}{l sin \theta}$ (2)
Divide (2) by (1).
$tan \theta = \frac{v^{2}}{gl sin \theta} \\\\ v^{2} = gl\, sin \theta \,tan \theta \\\\ v= \sqrt{gl \, sin \theta \, tan \theta}$

Answer: $v = \sqrt{gl \, sin \theta \, tan \theta}$

5 0

1 year ago

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