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1 hour ago

An object pulled to the right by two forces has an acceleration of 2.5 m/s2. The free-body diagram shows the forces acting on th

e object.
A free body diagram with 4 force vectors. The first vector is pointing downward labeled F Subscript g Baseline, the second pointing up labeled F Subscript N Baseline. The third pointing right labeled F Subscript 1 Baseline = 50 N the last is from the tip of the third using the tail to tip method labeled F Subscript 2 Baseline = 75 N pointing right.

What is the weight of the object?

50 N
98 N
313 N
490 N

Physics

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Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The

Keith_Richards [3271]

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m$

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

$\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s$

The person's speed is 11.56066 m/s

3 0

3 months ago

A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of t

serg [3582]

Response:

A protractor to gauge the angle between the inclined plane and the horizontal

Explanation:

The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

$f=\mu\,* N = \mu * m g\, cos(\theta)$

and $w_{//}= m\,g\,sin(\theta)$

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

$f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)$

For this, the sole additional tool needed is a protractor for angle measurement.

7 0

2 months ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Ostrovityanka [3204]

This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
To solve, utilize F=(K*Q1*Q2)/r^2 
You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C

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2 months ago

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Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h

kicyunya [3294]

I do not agree. Many materials may fluoresce when exposed to ULTRAVIOLET light, not in microwaves.:)

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2 months ago

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