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3 years ago

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You're using a wedge to split a log. you are hitting the wedge with a large hammer to drive it into the log. it takes a force of

1900 n to push the wedge into the wood. if the wedge moves 0.23 m into the log, how much work have you done on the wedge?

2 answers:

KIM [24]3 years ago

6 0

The work performed on an object is the force multiplied by the distance it is moved, provided the movement is parallel to the force. Since that is the case here, we can get the work by W=Fd=1900N x 0.23m = 437J. This energy is used to split the wood.

RoseWind [281]3 years ago

5 0

Answer: 437 J

Work done can be defined as the product of force and displacement. It is the work required to move an object to certain distance using a certain force.

$work=force\times displacement$

It is given that:

$Force=1900 N$

$Displacement=0.23 m$

Then, $work=1900N\times 0.23m = 437 N-m=437J$

Hence, the work done on the wedge to drive it into log is 437 J

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Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

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ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

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t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

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2 years ago

Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.

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3 years ago

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

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3 years ago

What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?

kvv77 [185]

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

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Explanation:

if

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

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$W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2$

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Juliette [100K]

Answer:

It takes 77 N

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