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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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According to Darwin's theory of evolution, how does natural selection cause species to evolve over time?
Elena L [17]

As time goes on, members of a population with favorable traits continue to breed while those without these traits eventually die off. For instance, if all of the food in an area was high up in the trees, only members of the species with long necks are able to eat and continue to live and reproduce. Those with shorter necks will not be able to eat and will eventually die off. Over time, the species will evolve to have long necks.

Hope this helps!

4 0
3 years ago
Two 1.1 kg masses are 1 m apart (center to center) on a frictionless table. Each has +10 JC of charge. What is the initial accel
sweet-ann [11.9K]

Answer:

acceleration = 0.8181 m/s²

Explanation:

given data

mass = 1.1 kg

apart d = 1 m

charge q = 10 μC

to find out

What is the initial acceleration

solution

we know that acceleration is

acceleration = \frac{force}{mass}   .................1

here force = k \frac{q1q2}{r^2}

here q1 q2 is charge and r is distance and Coulomb constant k = 9 × 10^{9} Nm²/C²

force = 9*10^{9} \frac{(10*10^{-6})^2}{1^2}

force = 0.9 N

so  from equation 1

acceleration = \frac{0.9}{1.1}

acceleration = 0.8181 m/s²

6 0
3 years ago
A plane flies 600 kilometers away from its bad at 400 meters per second then flies back to its base at 600 meters per second. Wh
tester [92]
Seems to me that it flies 400 m/s there and 600 m/s back the same distance.
therefore the average of 400 and 600 is 500 m/s.  The distance is the same so the normal formula of (d2-d1)/(t2-t1) is applicable.
4 0
3 years ago
Which law of motion accounts for the following statement?
m_a_m_a [10]

First, you need forces attracting the Sun and Earth toward each other.  That's described by Newton's law of universal gravitation.

Then you need the description of how that force causes the planet to move in a circular path.  The best choice is Newton's <em>Second law</em> of motion: F = m A .

4 0
3 years ago
2.
gizmo_the_mogwai [7]

Answer:

a) P1=100kpa

V1=6m³

V2=?

P2=50kpa

rearranging mathematically the expression for Boyle's law

V2=(P1V1)/P2=(100×6)/50=12m³

b) same apartment as in (a) but only the value of P2 changes

=> V2=(100×6)/40=15m³

Explanation:

since temperature is not changing we use Boyle's law. mathematically expressed as P1V1=P2V2

4 0
3 years ago
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