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netineya
4 days ago
6

A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a

t the rate of 1000 J/min. How long after starting will it take before the temperature starts to rise above 0°C
Physics
1 answer:
Maru [2.9K]4 days ago
4 0
The response is 176 minutes. The translation of 456 g equals 0.456 kg. The specific heat of ice is 2093 J kg⁻¹, used to calculate heat required for a 25-degree rise, determined by mass multiplied by specific heat and temperature increase. The necessary calculations yield a total heat load of 176164 J. Finally, by dividing heat required by heat supply rate, we ascertain that it will take approximately 176.16 minutes.
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1. Certainty in Reasoning: Douglas is confident in his ability to reason well to make sound judgments.

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1 month ago
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1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.
Keith_Richards [2897]

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

8 0
27 days ago
An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
Sav [2826]

Refer to the attached file for the solution

7 0
1 month ago
In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
kicyunya [2911]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak mg-T=\frac{mv^2}{r}

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as mg-0=\frac{mv^2}{r}

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top E_T=mg\times 2L+\frac{mv^2}{2}

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By comparing the two states using conservation of energy, we find v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

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6 0
26 days ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3084]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

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and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
1 month ago
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