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mixer [17]
3 years ago
5

A cheetah can run at approximately 100 km/hr and a gazelle at 80.0 km/hr. If both animals are running at fullspeed, with the gaz

elle 70.0 m ahead, how long before the che
Physics
1 answer:
brilliants [131]3 years ago
6 0

Answer:

12.6 seconds

Explanation:

With respect to the gazelle, the cheetah has a relative speed of 100 - 80 = 20 km/hr.

We convert from km/hr to m/s.

\dfrac{20000\text{ m}}{3600\text{ s}} = \dfrac{50}{9}\text{ m/s}

At this relative speed, the cheetah will overtake the gazelle in time <em>t</em> given by

v= \dfrac{d}{t}

<em>d</em> is the distance and <em>v</em> is the relative speed.

t = \dfrac{d}{v} = \dfrac{70.0}{50/9} = 12.6\text{ s}

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The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


4 0
3 years ago
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gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

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So

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and

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So the unit is

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2 years ago
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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo
Helga [31]

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

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A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786
ki77a [65]

Answer:

Explanation:

The explanation is given in the attached document.

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