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2 years ago

Tarzan swings back and forth on a long vine with a period of 7.27 s. how long is the vine?(unit=m)

Physics

1 answer:

lana [24]2 years ago

3 0

Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

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A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i

beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0

2 years ago

Read 2 more answers

The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0

2 years ago

If the mass of a material is 45 grams and the volume of the material is 8 cm^3, what would the density of the material be?

Svet_ta [14]

The density of the substance is the ratio of its mass over the space it occupies. In mathematical equation, this can be expressed as,

ρ = m / v

where ρ is density, m is mass, and v is volume.

Substituting the known values from the given,

ρ = (45 g) / (8 cm³)

ρ = 5.625 g/cm³

<em>ANSWER: 5.625 g/cm³</em>

3 0

2 years ago

a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

marissa [1.9K]

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

3 0

2 years ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$