The answer would be 79J because when you multiply 8.1*0.99*9.8 you get 7.89, which rounds to 7.9.
Answer choice (B).
a. True
b. False
situation is shown below. Blake makes the following claim about the free-body diagram:
Blake: “The velocity of the block is constant, so the net force exerted on the block must be zero.
Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied
force Fpull.”
What, if anything, is wrong with this statement? If something is
wrong, identify it and explain how to correct it. If this statement is
correct, explain why.
Answer:
The wrong items are;
1) The normal for FN equals the weight Fmg
2) The force of friction, Ff, equals the applied force Fpull
The corrected statements are;
1) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ)
Explanation:
The given statement was;
The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull
By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;
The normal force is equal to the weight less the vertical component of the applied force;
That is we have, FN = Fmg - Fpull × sin(θ)
The friction force similarly, is equal to the horizontal component of the applied force;
Ff = Fpull × cos(θ)
The wrong items are therefore as follows;
1) The normal for FN equals the weight Fmg
1 i) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction, Ff, equals the applied force Fpull
2 i) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ).
While Blake's statement about the normal force is correct, his claim about the applied force and friction force is partially accurate. In reality,the horizontal component of the applied force should equate to the friction force for the block to maintain a constant velocity.
Blake's claim that the normal force FN equals the weight Fmg is correct as these forces balance each other in the vertical direction. However, his claim that the force of friction Ff equals the applied force Fpull is only partially accurate. In reality, the horizontal component of Fpull (i.e., Fpull * cos(θ)) should equate to the friction force Ff, to maintain the constant velocity (the block is not accelerating). The vertical component of Fpull (i.e., Fpull * sin(θ)) reduces the effective weight of the block and thereby, the normal force.
To correct Blake's claim, the normal force FN is equal to the weight of the block minus the vertical component of the applied force, and the applied force's horizontal component equals the friction force. Hence, this is the correct solution considering both vertical and horizontal components of forces.
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Answer:
More powerful
Explanation:
I answered it and got it right.
The change in velocity of a car that starts at rest and has a final velocity of 20 meters/second north would be 20 meters/second.
The rate of change of the velocity with respect to time is known as the acceleration of the object. Generally, the unit of acceleration is considered as meter/seconds².
As given in the problem we have to find the change in velocity of a car that starts at rest and has a final velocity of 20m/s north,
the final velocity of the car = 20 meters/second north
Change in the velocity = final velocity - initial velocity
=20 - 0
= 20 meter/second
Thus, the change in the velocity would be 20 meters/second
Learn more about acceleration from here
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Generally, as temperature rises, solid solubility increases while gas solubility decreases. This is why when a carbonated drink gets warm, it has less bubbles.
Generally, as the temperature increases, the solubility of solids typically increases, while the solubility of gases usually decreases. Thus, for Ava's case with the beverage, as the drink got warmer (indicating an increase in temperature), the bubbles, which are gaseous, decreased. A real-life example of this is a carbonated drink. When you open a cold soda, it starts out very bubbly.
However, if you let it sit and warm up, the gas (carbon dioxide) becomes less soluble in the liquid and leaves the solution, resulting in fewer bubbles.
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