Answer:
Pressure of liquid depends on the shape of its container. This statement is true. The following reasons are mentioned below to justify the statement-
1. Particles in liquids are constantly and haphazardly streaming every which way.
2. As the particles advance, they continue to slam into one another.
3. The pressure applied by the liquids is similar every which way as per Pascal's rule.
4. The pressure at a spot inside a liquid is straightforwardly proportional to its depth, density, and gravitational acceleration.
5. It is free of the shape and size of the liquid-containing container.
Subsequently, the given statement is true.
A proton traveling at 3.60m/s suddenly enters a uniform magnetic field 0.750 T, traveling at an angle of 55 degrees.
a) Find the magnitude and direction of the force this magnetic field exerts on the proton.
b) If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve, and show how the velocity should be oriented to achieve these forces.
c)What would the answers to part (a) be if the proton were replaced by an electron traveling in the same way as the proton?
The force on an electron would be upwards. To achieve this force, the velocity of the proton should be oriented parallel to the magnetic field, as shown below:
What direction do the E and B fields follow?A magnetic field with north and south poles, similar to a bar magnet, is created as current flows through the coil. The magnetic field inside the solenoid has a parallel, linear pattern. A solenoid's magnetic field moves from its north pole to its south pole in one direction only.
a) To find the magnitude of the force on the proton, we use the formula:
F = qvBsinθ
where F is the force, q is the charge of the proton, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Plugging in the values given, we get:
[tex]F = (1.602 * 10^-^1^9 C) * (3.60 m/s) * (0.750 T) * sin(55) = 2.06 * 10^-^1^7 N[/tex]
To find the direction of the force, we use the right-hand rule, which states that if you point your thumb in the direction of the velocity, and your fingers in the direction of the magnetic field, then the force on a positive charge will be perpendicular to both your thumb and your fingers, in a direction given by the palm of your hand. In this case, the force will be pointing downwards, as shown below:
b) The maximum force will be achieved when the velocity of the proton is perpendicular to the magnetic field, since sin(90°) = 1, and the force is proportional to sinθ. In this case, the force will be:
[tex]F_m_a_x = (1.602 * 10^-^1^9 C) * (3.60 m/s) * (0.750 T) x sin(90) = 3.22 * 10^-^1^7 N[/tex]
To achieve this force, the velocity of the proton should be oriented at a right angle to the magnetic field, as shown below:
The minimum force will be achieved when the velocity of the proton is parallel to the magnetic field, since sin(0°) = 0, and the force is proportional to sinθ. In this case, the force will be zero:
[tex]F_m_i_n = (1.602 * 10^-^1^9 C) * (3.60 m/s) x (0.750 T) * sin(0) = 0 N[/tex]
To achieve this force, the velocity of the proton should be oriented parallel to the magnetic field, as shown below:
c) For an electron traveling in the same way as the proton, the magnitude of the force would be the same as in part (a), since the only difference between a proton and an electron is their charge (protons are positive, while electrons are negative). However, the direction of the force would be opposite, since electrons have a negative charge. Therefore, the force on an electron would be upwards.
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Using the Left Hand Rule, if motion is to the left and the field points up, which way is the current?
A. Toward you
B. Right
C. Away from you
D. Left
EXPLANATION:
The Left Hand Rule is a mnemonic device used to determine the direction of the magnetic field, current, or force in a conductor. The thumb, forefinger, and middle finger of the left hand are held perpendicular to each other, with the forefinger pointing in the direction of the magnetic field, the middle finger in the direction of the current, and the thumb in the direction of the force.In this case, if the motion is to the left, then the current must be flowing in the opposite direction, which is to the right. If the field points up, then the forefinger points up, and the middle finger must point to the right to satisfy the left hand rule. Therefore, the current is flowing towards you.
A 21 g block of ice is cooled to −77 ◦C. It is added to 593 g of water in an 92 g copper calorimeter at a temperature of 28◦C. Find the final temperature. The specific heat of copper is 387 J/kg · ◦C and of ice is 2090 J/kg · ◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg · ◦C . Answer in units of ◦C. ( 2 significant digits pls)
The final temperature of block when 593 g of water in an 92 g copper calorimeter at a temperature of 28°C is 27.96 °C.
First, we need to determine the amount of heat required to cool the ice to -77°C, then the amount of heat needed to melt the ice at -77°C, and finally the amount of heat needed to raise the temperature of the melted ice and the water to the final temperature.
The amount of heat required to cool the ice can be calculated using:
Q1 = m₁ * c₁ * ΔT₁
where m₁ is the mass of the ice, c₁ is the specific heat of ice, and ΔT₁ is the change in temperature.
Q₁ = 0.021 kg * 2090 J/kg·°C * (-77°C - 0°C) = -33.786 J
The negative sign indicates that heat is leaving the system.
The amount of heat required to melt the ice can be calculated using:
Q₂ = m₁ * L
where L is the latent heat of fusion of water.
Q₂ = 0.021 kg * 3.33 × 105 J/kg = 6993 J
The amount of heat needed to raise the temperature of the melted ice and water to the final temperature can be calculated using:
Q₃ = (m₂ * c₂ + m₃ * c₃) * ΔT₃
where m₂ is the mass of the copper calorimeter, c₂ is the specific heat of copper, m₃ is the mass of the water, c₃ is the specific heat of water, and ΔT₃ is the change in temperature.
Q₃ = (0.092 kg * 387 J/kg·°C + 0.593 kg * 4186 J/kg·°C) * (T - 28°C)
Simplifying and substituting the known values:
Q₃ = (35.604 + 2482.738) * (T - 28°C) = 88.7576 * (T - 28°C)
Now we can set up the equation to find the final temperature:
Q₁ + Q₂ + Q₃ = 0
-33.786 J + 6993 J + 88.7576 * (T - 28°C) = 0
Solving for T:
T = 27.96 °C
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38
A
B
2 weeks. Her data are summarized in the histogram.
Emily goes on walks through her neighborhood each day. She records data from each walk for
C
D
5
4
Number of Walks
What does the histogram show?
the number of walks within specific ranges of distances
the total number of kilometers walked for the 2 weeks
EMILY'S WALKS OVER TWO WEEKS
0 0-0.9 1.0-1.9 2.0-2.9 3.0-3.9 4.0-4.9 5.0-5.9 6.0-6.9
Distance (kilometers)
the number of walks within specific ranges of time lengths
the speed, in kilometers per hour, walked on different days
The histogram shown represents the number of walks that Emily went on during a two-week period, categorized by the distance she walked each day.
The x-axis of the histogram displays different distance ranges, from 0 to 6.9 kilometers, while the y-axis represents the frequency or number of walks that Emily went on in each distance range. The histogram indicates that Emily went on the most walks that were less than 0.9 kilometers and the frequency of walks decreased as the distance range increased.
Based on the information provided in the histogram, we cannot determine the total number of kilometers walked by Emily in the two-week period or her walking speed on different days. We can only infer that she went on more walks that were shorter in distance.
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