The total heat energy required to convert 76.0 g of ice at -18.0°C to water at 25.0°C is 35.184 kJ. This includes the energy to heat the ice to 0°C, melt the ice, and heat the water to 25°C.
Explanation:To calculate the heat energy required, we need to account for three processes: heating the ice to 0°C, melting the ice, and then heating the water to 25°C.
For the first process, we use the formula Q=mcΔT, where m is mass, c is specific heat, and ΔT is the temperature change. Ice has a specific heat of 2.09 J/g°C. So Q = 76.0 g * 2.09 J/g°C * 18°C = 2854.56 J.
For the second part, we use the formula Q = mLf, where m is mass and Lf is heat of fusion. For ice, Lf = 334 kJ/kg or 334 J/g. So Q = 76.0 g * 334 J/g = 25384 J.
For the last process, we once again use Q=mcΔT, this time with the specific heat of water, 4.184 J/g°C. So Q = 76.0 g * 4.184 J/g°C * 25°C = 7946 J.
To get the total heat energy required, we add these three quantities together and convert from joules to kilojoules. Q_total = (2854.56 J + 25384 J + 7946 J) / 1000 = 35.184 kJ. So, it requires 35.184 kJ of heat energy to convert 76.0 g of ice at -18.0°C to water at 25.0°C.
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To get an spaceship into orbit, it has to move
about as fast as the speed of sound
about as fast as the speed of light
about 18,000 miles per hour
there is no minimum speed for orbital motion
A negatively charged object is located in a region of space where the electric field is uniform and points due north. the object may move a set distance d to the north, east, or south. rank the three possible movements by the change in electric potential energy (ue) of the object. rank from greatest increase to decrease in ue.
(a) There will be an increase in the potential energy when the charge moves North.
(b) The variation of potential energy will be zero when the charge is moving to the east.
(c)There is a loss in potential energy when a charge moves to the South.
What will be the potential energy of the charge at different directions?(a) The largest increase in potential energy occurs when the charge is moving north. This is because the charge is negative,
so it acquires potential energy when moving in the same direction of the field (vice versa, a positive charge when moving in the direction of the field loses potential energy converting it into kinetic energy).
The amount of potential energy gained is equal to the product of the charge and the distance covered:
[tex]\rm \Delta PE=qD[/tex]
(b) The second-largest increase is when the charge is moving east. In this case, actually, the variation of potential energy is zero.
This is because the charge is moving perpendicular to the field, and so it is moving along points with the same potential. Therefore, in this case, the variation of potential energy is zero:
[tex]\rm \Delta PE=0[/tex]
(c) Finally, when the charge is moving south, it loses potential energy. This is because it is moving against the electric field,
since it is a negative charge, in this direction it loses potential energy converting it into kinetic energy. Therefore, in this case
[tex]\rm \Delta PE=-qD[/tex]
Thus
(a) There will be an increase in the potential energy when the charge moves North.
(b) The variation of potential energy will be zero when the charge is moving to the east.
(c)There is a loss in potential energy when a charge moves to the South.
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Without inertia, how would an object that is experiencing a centripetal force behave?
It would move in a line away from the circle’s center.
It would move in a line toward the circle’s center.
It would move in a curved, circular path.
It would move in a line tangent to the circular path.
B. It would move in a line toward the circle’s center.
Explanation:
got it right on edge 2022
Brad is working on a speed problem in physics class. The problem tells him that a girl runs from her house to the park 0.05 km away in 10 s. Brad calculates that her speed is 0.005 m/s. Is he correct? If not, explain the flaw or flaws in his problem solving process.
Answer:
He is incorrect! Her speed was 5m/s.
Explanation:
For calculating the speed, first we shall remember that:
[tex]v=\dfrac{d}{t}[/tex]
Where [tex]v[/tex] is the speed, [tex]d[/tex] is the distance travelled and, [tex]t[/tex] is the time it takes to travel distance [tex]d[/tex].
So one migth think that velocity can be easely compute:
[tex]v=\dfrac{0.05}{10}[/tex]
[tex]v=0.005\dfrac{m}{s}[/tex]
Be carefull, he does not make a proper dimensional analisis!
Before computing the speed we must know in what dimensions our values are.
[tex]d=0.05km[/tex], distances is measure in Kilometers.
[tex]t=10s[/tex], time is measure in seconds.
If we want our speed to be in [tex]m/s[/tex], first we need to be sure that our values are expressed in meters and seconds.
Time is already expressed in seconds, distance is not in Kilometers.
So
[tex]0.05Km=50m[/tex],
now we can compute the speed:
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{50m}{10s}[/tex]
[tex]v=\5dfrac{m}{s}[/tex]
You walk into an elevator, step onto a scale, and push the "up" button. you recall that your normal weight is 639 n . when the elevator has an upward acceleration of magnitude 2.90 m/s2 , what does the scale read?
This question deals with the concepts of the actual weight and apparent weight.
The apparent weight of the person is "827.9 N".
APPARENT WEIGHTThe apparent weight of an object is the reaction of the elevator floor on the person while the elevator is in accelerated motion. It is not the actual weight but the weight felt by the person for that time. In this case the elevator is moving up. Hence the apparent weight will be:
[tex]W_a=m(g+a)=mg+ma\\W_a=W+ma[/tex]
where,
W = actual weight = 639 Nm = mass = [tex]\frac{W}{g}=\frac{639\ N}{9.81\ m/s^2}[/tex] = 65.14 kga = acceleration = 2.9 m/s²[tex]W_a[/tex] = apparent weight = ?Therefore,
[tex]W_a=639\ N + (65.14\ kg)(2.9\ m/s^s)[/tex]
[tex]W_a=827.9\ N[/tex]
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The scale reading in an elevator accelerating upwards will display an increased weight due to the additional force of acceleration. When an elevator accelerates with a magnitude of 2.90 m/s^2, the scale will show a higher value than the normal weight, calculated by the sum of gravitational force and force of acceleration.
Explanation:When you step onto a scale in an elevator that is accelerating upwards with a magnitude of 2.90 m/s2, the scale reading will be higher than your normal weight due to the additional force required to accelerate you upwards. Given that your normal weight is 639 N, we can calculate the new scale reading by incorporating the effects of the elevator's acceleration using Newton's second law of motion.
To find the new scale reading, we first determine the apparent weight. The apparent weight is the sum of the true weight (gravitational force) and the force of acceleration (ma).
Apparent weight = True weight (W) + Force of acceleration (ma)
Where the true weight W = mg (mass times gravity), and a is the acceleration of the elevator.
Assuming Earth's gravity to be 9.81 m/s2, we can calculate the apparent weight as follows:
Apparent weight = mg + ma
Now, we need to find the mass (m) from the given weight (639 N), which is m = W/g = 639 N / 9.81 m/s2.
Then plug the mass and the given acceleration into the equation for apparent weight.
The scale reading in an accelerating elevator is directly proportional to the acceleration; it increases as the elevator accelerates upwards. However, once the elevator reaches a constant velocity, the scale reading will return to your normal weight, 639 N, because there will be no additional force from acceleration (a = 0).
How many cells must be connected in series to give the 350 v a large catfish can produce?
Final answer:
Approximately 2333 electro plaques, or biological cells, would be needed to be connected in series to produce 350 V, given that each cell produces 0.15 V.
Explanation:
To calculate the number of cells required to produce a voltage of 350 V, similar to what a large catfish can produce, we must understand that when cells are connected in series, the total voltage is the sum of the individual voltages of each cell. If each cell, similar to those in an electric eel, produces an electromotive force (emf) of 0.15 V, then the number of cells required to reach 350 V would be the total voltage desired divided by the voltage of one cell.
(350 V) / (0.15 V per cell) = 2333.33 cells
Therefore, approximately 2333 cells would need to be connected in series to produce 350 V. This figure is derived by understanding that the potential difference across each cell adds up when they are connected in a series circuit, a principle that is crucial in the functioning of biological cells called electro plaques in electric fish.