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3 years ago

Tarzan, in one tree, sights Jane in another tree

2 answers:

8 0

Taking the distance of Tarzan from the ground before and after he makes the swing:

Ho (initial height) = L(1 - cos45) = 20 (1 - 0.707) = 5.86 meters
Hf (final height) = L(1 - cos30) = 20 (1 - 0.866) = 2.68 meters

Difference in height = 5.86 - 2.68 = 3.18 meters

PE = KE
mgh = (1/2)mv^2

Solving for v:
v = sqrt (2*g*h)
v = sqrt (2*9.8*3.18)
v = 7.89 m/s

With Tarzan going that fast, it is likely that he will knock Jane off.

kobusy [5.1K]3 years ago

8 0

Answer:

Tarzan's speed is 7.91m/s.

Tarzan will knock off Janefrom the tree.

Explanation:

Initial vertical height = L-LcosA = L(1-cos45)

Initial vertical height = 20× 0.29= 5.8m

Final vertical height = L(1-cos30°)= 20× 0.13

Final vertical height = 2.6m

Loss in vertical and = 5.8- 2.6=3.2m

Loss in Potential energy = Gain in kinetic energy

Mgh=1/2(mv^2)

V= sqrt( 2× 9.8 × 3.2)

V= 7.9m/s

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Answer:

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Frequency will be equal to $12\times 10^8Hz$

Explanation:

We have given frequency of the radar f = 10 GHz $=10\times 10^6Hz$

Speed of light $c=3\times 10^8m/sec$

Plank's constant $h=6.6\times 10^{-34}js$

So energy $E=h\nu$ ,here h is plank's constant and $\nu$ is frequency

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In second case we have given wavelength = 25 cm = 0.25 m

Wavelength is equal to $\lambda =\frac{c}{f}$

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3 years ago

A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

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3 years ago

A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho

Lemur [1.5K]

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

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3 years ago

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

From first equation of motion we know that

v = u+at

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So final velocity will be -9.28 m/sec

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3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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