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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<span>) = 2.68 meters
</span>
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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Un soldado de 1.80 m de altura ha sido herido por una bala, cuya masa es de 100 g. Deciden colocar al soldado en una centrifugad

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3 years ago

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

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6 0

1 year ago

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s

Calculation of the acceleration of the hockey puck

We apply the Formula (1)

vf=v₀+a*t v₀=5.8 m/s , vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the frictional force (Ff)

We apply Newton's second law

∑F = m*a Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the hockey puck in the attached graphic

∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²

-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N = 1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0

2 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$