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Rasek [7]
3 years ago
10

The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is

13 cm and the width is 5 cm, how fast is the area of the rectangle increasing?
Physics
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

24cm/s

Explanation:

A=L*w

A'=L'*w'

L=13

w=5

L'=4

w'=6

A=?

A'=?

A=L*w

A=13*5

A=65

A'=L'*w'

A'=4*6

A'=24

*the given lengths are just to throw you off*

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8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in
Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N.  If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.

8 0
3 years ago
A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released
GREYUIT [131]

Answer:

6.75\mu C/m^2

Explanation:

We are given that

Diameter,d=1\mu m=1\time 10^{-6} m

1\mu m=10^{-6} m

Radius,r=\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m

Density,\rho=900kg/m^3

Total number of electrons,n=39

Charge on electron =1.6\times 10^{-19} C

Total charge=q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C

Distance,s=2mm=2\times 10^{-3} m

Mass =density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg

Initial velocity,u=0

Final speed,v=4.5 m/s

v^2-u^2=2as

(4.5)^2-0=2a(2\times 10^{-3})

20.25=4a\times 10^{-3}

a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2

Force,F=ma

qE=ma

q(\frac{\sigma}{2\epsilon_0})=ma

\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}

\epsilon_0=8.85\times 10^{-12}

\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2

6 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
adoni [48]

Answer:

B) R1 = 6 V and R2 = 6V

Explanation:

In series, both resistors will carry the same current.

that current will be I = V/R = 12 / (10 + 10) = 0.6 A

The voltage drop across each resistor is V = IR = 0.6(10) = 6 V

6 0
3 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
3 years ago
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Sindrei [870]

Answer: _____= beautiful, yet annoying frustrating death

Explanation:

3 0
3 years ago
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