An object is thrown with an initial speed v near the surface of Earth. Assume that air resistance is negligible and the gravitat
ional field is constant. An object is thrown with an initial speed u near the surface of Earth. Assume that air resistance is negligible and the gravitational field is constant. If the object is thrown horizontally, the direction and magnitude of its acceleration while it is in the air is _________.a. upward and decreasing b. upward and constant c. downward and decreasing d. downward and increasing e. downward and constant
2 answers:
Answer:
E. downward and constant
Explanation:
Freefall is a special case of motion with constant acceleration because the acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.
For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.
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The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
Answer:
Show attached picture
Explanation:
Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call its internal resistance) and R indicates the resistance of the light bulb.
We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:
(1)
Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:
Using Ohm's law, , we can rewrite the previous equation as:
where
is the current in the meter
is the current in the bulb
Using (1), this equation becomes
so, the current in the meter is 1000 times less than through the bulb.
