The greatest acceleration that the truck can have before the toolbox slides out can be calculated by understanding the balance between the inertia force experienced by the toolbox due to acceleration (F = ma) and the maximum static friction force (fs(max) = μsN) opposing this motion. The truck can accelerate up to the point at which these two forces are equal.
The question relates to a concept in Physics known as Friction. In this scenario, the toolbox on the truck experiences static friction which keeps it from sliding. The maximum force of static friction can be calculated using the equation fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. In this case, μs is given as 0.300 and the normal force N equals the weight of the toolbox. The truck can accelerate up to the point where the frictional force equals the force caused by acceleration, which is calculated using the equation F = ma, where m is mass and a is acceleration.
When the truck accelerates, an inertia force acts on the toolbox in the opposite direction. This inertia force, F = ma, should not exceed the maximum static friction force, fs(max), otherwise, the toolbox will slide. Hence, with given values of static friction coefficient and mass of the toolbox, the greatest acceleration of the truck to prevent slipping can be calculated by equating the frictional force and inertia force.
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The greatest acceleration that the truck can have before the toolbox slides out is 5.00 m/s².
The greatest acceleration that the truck can have before the toolbox slides out can be found by comparing the force of static friction to the force pushing the toolbox forward. In this case, the force of static friction must be equal to or greater than the force pushing the toolbox, which is the product of the mass of the toolbox and its acceleration. Given the coefficient of static friction of 0.300, the maximum force of static friction can be calculated. Using the equation fs <= μsN, where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force, we can substitute the values and solve for the maximum force of static friction which is 196 N. The maximum force of static friction is equal to the product of the mass of the toolbox and its acceleration, which gives us the equation fs = max = (50.0 kg)(5.00 m/s²) = 250 N. Therefore, the greatest acceleration that the truck can have before the toolbox slides out is 5.00 m/s².
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The actual diameter of the heart is 12.25 cm. Given : Heart measure = 14.7 cm Magnification factor = 1.2
14.7 / 1.2 = 12.25
increase by a factor of 4
decrease by a factor of 2
decrease by a factor of 4
Answer:
Increase by a factor of 4.
Explanation:
The acceleration of a car moving with speed v in a circle of radius R is given by:
Now if we double the speed in the equation above, it becomes . Thus:
Therefore the acceleration is increased by a factor of 4.
Answer:
B 1.08 BILLION
Explanation:
SEE ATTACHMENT
Answer:
The total mass of the abattoir is 100 grams. When you add 100 ml of water and 50 ml of oil, you need to consider the density of these substances to calculate their mass.
The density of water is approximately 1 gram per milliliter (g/ml). So, 100 ml of water has a mass of 100 grams.
The density of oil can vary depending on the type, but for a rough estimate, we can assume it's around 0.9 grams per milliliter (g/ml). So, 50 ml of oil has a mass of 50 x 0.9 = 45 grams.
Now, you can calculate the total mass:
Total mass = Mass of abattoir + Mass of water + Mass of oil
Total mass = 100 grams + 100 grams + 45 grams
Total mass = 245 grams
The total mass when 100 ml of water and 50 ml of oil are added to the abattoir will be 245 grams.
Explanation:
The buoyancy of an object depends on its density relative to the density of the fluid it's placed in, typically water in everyday scenarios. When an object is placed in a fluid (like water), it will float if its average density is less than that of the fluid. In other words, if the object is less dense than the fluid, it will float.
In this case, both pieces of wood have the same density of 0.90 grams per milliliter (g/ml), which is less dense than water. Therefore, both the 1-pound piece of wood and the 10-pound piece of wood will float in water.
However, it's important to note that the buoyant force acting on these objects will be the same for a given volume of wood because they have the same density.
The 10-pound piece of wood will displace more water (have a larger volume) to support its weight, while the 1-pound piece of wood will displace less water (have a smaller volume) to support its weight. The 10-pound piece of wood will have more of its volume submerged compared to the 1-pound piece due to the weight difference, but both will float.