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Gnom [1K]
3 years ago
9

Tarzan, whose mass is 103 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets

go, his center of mass is at a height 2.4 m above the ground and the bottom of his dangling feet are at a height 1.5 above the ground. When he first hits the ground he has dropped a distance 1.5, so his center of mass is (2.4 - 1.5) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.3 above the ground.
Required:
Consider the real system. What is the net change in internal energy for Tarzan from just before his feet touch to the ground to when he is in the crouched position?
Physics
1 answer:
Anettt [7]3 years ago
6 0

Answer:

   v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

Explanation:

For this exercise let's start by using Newton's second law

Y axis

        N-W = 0

        N = W

X axis

         fr = m a

the expression for the friction force is

         fr = μ N

we substitute

        μ mg = m a

        μ g = a

calculate us

         a = 0.620  9.8

         a = 6.076 m / s²

now we can use the kinematics relations

          v² = v₀² - 2 a x

suppose v = 0

          v₀ = \sqrt{2ax}Ra 2ax

let's calculate

         v₀ = \sqrt{2 \ 6076 \ 60}

         v₀ = 27.00 m / s

let's slow down to the english system

          v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)

          v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

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A long solenoid has a diameter of 11.1 cm. When a current i exists in its windings, a uniform magnetic field of magnitude B = 42
Pepsi [2]

Answer:

E(1) = 1.44\times 10^{-4} v/m

E(2) = 1.34\times 10^{-4} v/m

Explanation:

Given data:

diameter of solenoid is 11.1 cm

B = 42.9 mT

dR = 6.65 mT/s

d(1) = 4.34 cm

d(2) = 7.46 cm

we know that

electric field due to solenoid is given as

E = 1/2 dB/dt (r)

E(1)= \frac{0.0434}{2} \times (6.65\times 10^{-3})

E(1) = 1.44\times 10^{-4} v/m

E = 1/2 dB/dt (R^2)/r

E(2)= \frac{0.055^2}{2\times 0.0746} \times (6.65\times 10^{-3})

E(2) = 1.34\times 10^{-4} v/m

6 0
3 years ago
"a 1,600 kg car is traveling at a speed of 12.5 m/s. what is the kinetic energy of the car?"
xxTIMURxx [149]
I had this question before can you show me the answers choices
6 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10−5 N · m2 /C when th
Zanzabum

Answer:

\sigma=2.124\times 10^{-13}C/m^{2}

Explanation:

Given:

Electric Flux = 3\times10^{-5}N.m^2/C

Side of sheet = 5cm

Area of the square sheet, A= 5×5 = 25cm²=25×10⁻⁴ m²

Now

the electric flux (Φ) is given as:

\phi =EA

where, E = Electric field

or

E=\frac{\phi}{A}

substituting the values in the above equation, we get

E=\frac{3\times10^{-5}N.m^2/C}{25\times 10^{-4}}

E=0.012N/C

Now the charge density (σ) on a sheet is given as:

\sigma=2\epsilon_oE

where, \epsilon_o = Permittivity of the free space = 8.85×10⁻¹²

substituting the values in the above equation, we get

\sigma=2\times 8.85\times10^{-12}\times 0.012

\sigma=2.124\times 10^{-13}C/m^{2}

5 0
3 years ago
Knowing the definition of work as used by physicists, how can the work done on an object be equal to zero? a. The force and dist
Nuetrik [128]

Answer:

C

Explanation:

Work is the product of force and distance moved in the direction of force . if no distance, no work is done. Holding a heavy object still means no work is done

4 0
3 years ago
An incident ray of light strikes a diamond at an angle of
lys-0071 [83]

Answer:

The angle of refraction is option b: 17°.

Explanation:

We can find the angle of refraction by using Snell's law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})

Where:

n₁: is the index of refraction of the medium 1 (air) = 1.0003  

n₂: is the index of refraction of the medium 2 (diamond) = 2.42

θ₁: is the angle of incidence = 45°

θ₂: is the angle of refraction =?

Hence, the angle of refraction is:

sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{1.0003*sin(45)}{2.42} = 0.2922

\theta_{2} = 17 ^{\circ}

Therefore, the correct option is b: 17°.

I hope it helps you!

4 0
3 years ago
Read 2 more answers
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