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Softa [21]
3 years ago
13

How many joules of heat must be transferred to a 410-g aluminum pizza pan to raise its temperature from 32oC to 232oC? The speci

fic heat of aluminum in this temperature range is 0.96 J/g x oC.
Physics
1 answer:
xxTIMURxx [149]3 years ago
3 0

Answer:

recall that heat absorbed released is given by

Q = mc*(T2 - T1)

where

m = mass (in g)

c = specific heat capacity (in J/g-k)

T = temperature (in C or K)

*note: Q is (+) when heat is absorbed and (-) when heat is released.

substituting,

Q = (480)*(0.97)*(234 - 22)

Q = 98707 J = 98.7 kJ

Explanation:

You might be interested in
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.001
Vlad1618 [11]

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

3 0
3 years ago
Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0
3 years ago
Use the diagram to answer the question. What happens at night?
melamori03 [73]
The return flow goes to the right and instead of sea breeze, there's land breeze.
7 0
3 years ago
Read 2 more answers
A ball of gas becomes a ____ when nuclear fusion begins in its core.
Ira Lisetskai [31]
The answer to this is Protostar.

This is a process where it is gathering mass from its parent molecular cloud. Its a very young star meaning, the star was now born. 

Hope this helped :)
Have a great day
8 0
3 years ago
A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.
olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0
3 years ago
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