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3 years ago

9

a climber is attached to a vertical rope that has a tension of 1300 n. if the climber has a mass of 68 kg, what is the net force

on the climber

2 answers:

solmaris [256]3 years ago

6 0

634 N up is the answer

Anna71 [15]3 years ago

5 0

Answer:

634 N up

Explanation:

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Ket [755]

Answer:

1) D

2) A

Explanation:

1) Each group has the same number of valence electrons, which are the outer electrons.

2) Ionic bonds are between a metal and non - metal, the metal being sodium and the non - metal being chlorine.

3 0

3 years ago

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

4 0

3 years ago

Science: Which statement is true about star A

Elza [17]

Answer:

no clue

Explanation:

6 0

2 years ago

horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes

lbvjy [14]

Answer:

The tension is $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

The distance between the two poles is $D =14 m$

The mass tied between the two cloth line is $m = 3Kg$

The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium

And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

$= tan^{-1} [\frac{1}{7}]$

$=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

5 0

3 years ago

Read 2 more answers

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

3 0

3 years ago