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1.6.53m/s B.4.96m/s C. 1.48 m/s D.3.48m is IV 2 2 kg 3 kg ) of ice...
How much ice is in the final state after 2 kg of ice at -30°C is...
How much ice is in the final state after 2 kg of ice at -30°C is added to a large quantity of water at 0°C? The final state is ice-water. (The temperature of the ice increases to 0°C and some of the water freezes.) J For water : Cice = 4186 J kg.K If = 3 = 2100 kg.K : Cwater L, = 2.26 x 106 .34 x 10% J kg 9 kg
(20%) Problem 3: A thermos contains m = 0. 79 kg of tea at IT, = 31° C. Ice (m, = 0.055 kg, T, = 0° C) is added to it....
(20%) Problem 3: A thermos contains m = 0. 79 kg of tea at IT, = 31° C. Ice (m, = 0.055 kg, T, = 0° C) is added to it. The heat capacity of both water and tea is c 4186 J/(kg K), and the latent heat of fusion for water is L4= 33.5 x 104 J/kg. A 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal...
The temperature of 2.26 kg of water is 34 °C. To cool the water, ice at...
The temperature of 2.26 kg of water is 34 °C. To cool the water, ice at 0 °C is added to it. The desired final temperature of the water is 11 °C. The latent heat of fusion for water is 33.5 × 104 J/kg, and the specific heat capacity of water is 4186 J/(kg·C°). Ignoring the container and any heat lost or gained to or from the surroundings, determine how much mass m of ice should be added.
(20%) Problem 3: Athermos contains mj = 0.73 kg of tea at T 33° C. Ice...
(20%) Problem 3: Athermos contains mj = 0.73 kg of tea at T 33° C. Ice (m2 0.075 kg, T2 0° C) is added to it. The heat capacity of both water and tea is c 4186 J/(kg K), and the latent heat of fusion for water is L,= 33.5 x 104 Jkg. 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal equilibrium. Grade Summary T= Deductions 0%...
A bullet of horizontal velocity v1i = 400 m/s and mass m1 = 0.03 kg strikes,...
A bullet of horizontal velocity v1i = 400 m/s and mass m1 = 0.03
kg strikes, and passes through, a block of wood of mass m2 = 0.2 kg
which is initially at rest on a frictionless horizontal surface. If
the final velocity of the block is v2f = 25 m/s, what is the final
velocity of the bullet?
(7 pts) A bullet of horizontal velocity vi=400 m/s and mass m, = 0.03 kg strikes, and passes through a block...
A 0.014 kg bullet traveling horizontally at 403 m/s strikes a 4 kg block of wood...
A 0.014 kg bullet traveling horizontally at 403 m/s strikes a 4 kg block of wood sitting at the very edge of a horizontal table of height 1 meters. The bullet is lodged into the wood. [For the signs of directions in this problem, take forward and upward to be positive; backward and down to be negative.] a) What is the momentum of the bullet right before it impacts the block of wood? b) What is the momentum of the...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
How much ice is in the final state after 2 kg of ice at -30°C is added to a large quantity of water at 0°C? The final state is ice-water. (The temperature of the ice increases to 0°C and some of the water freezes.) J For water : Cice = 4186 J kg.K If = 3 = 2100 kg.K : Cwater L, = 2.26 x 106 .34 x 10% J kg 9 kg
(20%) Problem 3: A thermos contains m = 0. 79 kg of tea at IT, = 31° C. Ice (m, = 0.055 kg, T, = 0° C) is added to it. The heat capacity of both water and tea is c 4186 J/(kg K), and the latent heat of fusion for water is L4= 33.5 x 104 J/kg. A 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal...
(20%) Problem 3: Athermos contains mj = 0.73 kg of tea at T 33° C. Ice (m2 0.075 kg, T2 0° C) is added to it. The heat capacity of both water and tea is c 4186 J/(kg K), and the latent heat of fusion for water is L,= 33.5 x 104 Jkg. 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal equilibrium. Grade Summary T= Deductions 0%...
A bullet of horizontal velocity v1i = 400 m/s and mass m1 = 0.03
kg strikes, and passes through, a block of wood of mass m2 = 0.2 kg
which is initially at rest on a frictionless horizontal surface. If
the final velocity of the block is v2f = 25 m/s, what is the final
velocity of the bullet?
(7 pts) A bullet of horizontal velocity vi=400 m/s and mass m, = 0.03 kg strikes, and passes through a block...
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
The 0.02 kg bullet is travelling at 400 m/s when it becomes embedded in the 2 kg stationary block. The coefficient of kinetic friction between the block and plane is 0.2. A. Determine the velocity of the bullet and block just after the collision. B. Use the principle of linear impulse and momentum to determine the time during which the bullet/ block system will slide before it stops. 400 m/s
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