1 ton = 2240 lb 10 ton = 10 x 2240 lb = 22400 lb
8 oz + 9 lb 8 oz = 10 lb 2 lb 4 oz + 10 lb = 12 lb 4 oz 8 oz + 2 lb 4 oz = 2 lb 12 oz
no 10 times more . 1 lb = 16 oz 10 lb = 160 oz 100 lb = 1,600 oz
29 lb
The volume of the block of ice is 2ft * 2ft * 2ft or 8 ft3. The equivalent weight of 8 cubic feet of water is 8ft3 * 62.5 lb/ft3 or 500 lbs. To find the weight of ice take 92% of the equivalent weight of water or 0.92 * 500 lbs = 460 lbs
A block of ice with the dimensions listed has a volume of 35ft^3. The density of ice at 32F is about 57.24 lb/ft^3. Multiplying the two together gives a mass of the ice of about 2003.4 lbs.
1,2,3,4,5,6,7,8,9,10
I found that on a website selling a 379. BTW I'm not a trucker. DIMENSIONS GVW: 48,000 LB Front: 12,000 LB Rear: 36,000 LB W x H x L: 96" x 12' 10" x 29' 2" Wheelbase: 265" Cab to Axle: 83" from fairing 102" from sleeper (*Don't undestanding that..XD) Weight: Estimated: 18,000 LB
Depending on the mean temperature of the ice, it will require between 0.463 and 0.504 BTU's to raise the temperature of 1 lb. of ice for each deg. F. So, 5 lb of ice raised from 3 deg.F to 10 deg.F will require 5 x (10-3) x ~0.504 = ~17.64 BTU.
1 gallon of water weighs 8.34 lb so it will make 8.34 lb of ice
10% is equivalent to 1/10. 10% of 12000000 lb (1.2x107 lb) would be 1200000 lb (1.2x106 lb).
To calculate the number of 8 oz cups of ice in a 10 lb bag, you first convert the weight of the bag to ounces. Since 1 lb is equal to 16 oz, a 10 lb bag is equivalent to 160 oz. Next, you divide the total weight by the weight of each cup of ice (8 oz) to find the number of cups. Therefore, a 10 lb bag of ice would contain 20 cups (160 oz ÷ 8 oz = 20 cups).
The exact amount of cooling will depend on several factors such as the size of the pool, the initial temperature of the water, and the ambient temperature. In general, an 800 lb block of ice will cool a pool, but it may not be sufficient to significantly lower the pool's temperature, especially if the pool is large or the weather is warm. It's best to consider using additional ice or a different cooling method for more effective results.
The component of the force parallel to the incline is 30 lb * sin(20°) = 10.27 lb. The net force on the block up the incline is 10.27 lb - frictional force. The frictional force is equal to the block's weight component parallel to the incline, which is 60 lb * sin(20°) = 20.53 lb. So, the net force causing acceleration is 10.27 lb - 20.53 lb = -10.26 lb. Using F = ma, the acceleration a = (-10.26 lb) / (60 lb) = -0.17 ft/s^2 down the incline.
1 ton = 2240 lb 10 ton = 10 x 2240 lb = 22400 lb
18 stone and 9 lb.
1 pound