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solmaris [256]
3 years ago
13

A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on

top of it, then the spring is released. Approximately how high will the rock rise? 9 m 17 m 27 m 85 m
Physics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0

Answer:

9m

Explanation:

edge2o2o

Sedaia [141]3 years ago
5 0

Answer:

it's 9m

Explanation:

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.
Igoryamba

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

                                            = 25/ 8.5

                                             = 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

#SPJ4

5 0
1 year ago
Can anyone plz answer this now and give me a right answer?
klemol [59]

Answer:

90 degree hope it help :))

5 0
3 years ago
Read 2 more answers
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?​
Xelga [282]
<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

<h3>And,</h3>

30 cm = 0.3 m (distance)

<h3>So,</h3>

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

\small \boxed{work \: done \:  by \: Ryan \:  = 3 \: Joules}

4 0
3 years ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
3 years ago
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