What is the magnitude of a vector if the x component is - 12 m and the y component is 21m?290m
17m
24m
580m
Answers
Answer:
v = 24 m
Explanation:
Given that,
The x component of the vector = 12 m
The y component of the vector = 21 m
We need to find the magnitude of a vector. It can be solved as follows :
or
v = 24 m
So, the magnitude of the vector is equal to 24 m.
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B.
The object is at rest between points B and C because the distance has not changed.
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Answer:
In the lab activity, I used a virtual watershed to demonstrate how pollutants enter and flow through the watershed. I constructed food chains to show how biotic factors are related. I also predicted and modeled how human activity affected the biotic factors in an ecosystem. Based on my observations, I can conclude that biotic and abiotic factors are closely linked and that human activity that changes the environment, like pollution, can affect all the other parts of an ecosystem in a negative way.
Answer:
In the lab activity, I used a virtual watershed to demonstrate how pollutants enter and flow through the watershed. I constructed food chains to show how biotic factors are related. I also predicted and modeled how human activity affected the biotic factors in an ecosystem. Based on my observations, I can conclude that biotic and abiotic factors are closely linked and that human activity that changes the environment, like pollution, can affect all the other parts of an ecosystem in a negative way.
Explanation:
that exerts a force of magnitude F 12.0 N at an angle
u 25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before
it is lifted (completely) off the floor?
Answers
the force in the direction of movement
Fx = F x cosA*
Fx = 12 x cos25*
Fx = 10.88
m x a = 10.88
a = 10.88/5
a= 2.18 m/s^2
b.
when the block will be lifted off the floor and when the vertical component
Fy = mg
Fy = F x sin25*
Fy= 5 x 9.8
Fy= 115.94 N
c.
if Fx = ma
a = 115.94/5
a= 23.19 m/s^2
hope it helps
Final answer:
The magnitude of the block's acceleration is roughly 2.18 m/s². The force required just before the block is lifted off the floor is equal to the weight of the block, 49.05 N. Just before it is lifted, the block's acceleration is still due to the horizontal component of the force.
Explanation:
This problem involves physics concepts related to forces and acceleration of an object on a frictionless surface. The key to solving it involves understanding the relationship between mass, force, and acceleration (Newton's second law) and the concept of vertical and horizontal components of a force.
(a) The magnitude of the block's acceleration can be calculated using the horizontal component of the force (Fh = F cos θ) and Newton's second law (F = ma). So, a = Fh / m = (F cos θ) / m = (12.0 N cos 25) / 5.00 kg ≈ 2.18 m/s².
(b) The force F required to lift the block is equal to the weight of the block mg. Thus, F = m*g = 5.00 kg * 9.81 m/s² = 49.05 N.
(c) Just before the block is lifted off the floor, it is still in contact with the floor so the normal force is not zero. As such, there is no vertical acceleration, so the magnitude of the block's acceleration is still solely due to the horizontal component of F.
Learn more about Forces and Acceleration here:
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C.a mirror
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Answers
Answer:
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Explanation:
Took the test <3