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2 years ago

7

A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit

h the vertical. In such a condition of static equilibrium, what is the net force on the new vine? ANS: 366 N

1 answer:

zmey [24]2 years ago

4 0

Answer:

$T=366.23\ N$

Explanation:

Given:

mass of monkey, $w=600\ N$

angle of vine from the vertical, $\theta=35^{\circ}$

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

$\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}$

$\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }$

$T=366.23\ N$

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A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

Snezhnost [94]

We can solve the problem by using Snell's law, which states
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
From which we find
$\theta_r = \arcsin (0.288)=16.7^{\circ}$

6 0

2 years ago

Which of the choices describes an action–reaction force pair for a space station containing astronauts in orbit about the earth?

11111nata11111 [884]

Answer:

(D) The weight of the space station and the gravitational force of the space station on the earth.

Explanation:

In both A and B , both the forces act in the same direction ( downwards ) , so they can not be action- reaction force .

In the option C , weight of a astronaut can only be reaction force of gravitational force exerted on the earth by astronaut. Both astronaut and the earth pull each other with equal and opposite force. So option D is correct.

5 0

2 years ago

A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .

GREYUIT [131]

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

5 0

2 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

2 years ago

An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a ma

saul85 [17]

Answer:

98.15 lb

Explanation:

weight of plane (W) = 5,000 lb

velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s

wing area (A) = 200 ft^{2}

aspect ratio (AR) = 8.5

Oswald efficiency factor (E) = 0.93

density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

we need to get the drag coefficient (Cd) before we can solve for the drag

Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)

where

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ (take note that π is shown as n and ρ is shown as $p$ )

where lift coefficient (Cl)= $\frac{2W}{pAv^{2} }$ = $\frac{2x5000}{0.002377x200x293.3^{2} }$ = 0.245

therefore

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ = $\frac{0.245^{2} }{3.14x0.93x8.5}$ = 0.0024

since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)

Cd = 0.0024 + 0.0024 = 0.0048

Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb

8 0

2 years ago