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Strike441 [17]
3 years ago
9

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2 m/s2 for 6.5 s , after which the gazelle continues at a steady speed.
Physics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

v=at

Where, v = speed

a = acceleration

t = time

Put the value into the formula

v=4.2\times6.5

v=27.3\ m/s

Hence, The gazelles top speed is 27.3 m/s.

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S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
6 0
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Answer:

v=4.44\frac{m}{s}

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero)  and accelerates at a constant rate, position can be described like this: x=v_{0}t +\frac{1}{2} at^{2} where x is the position, t is the time a is the acceleration and v_{0} is initial velocity. In this way acceleration can be found. a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }.

Now we are able to found velocity at any time with the formula: v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}

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So i need help with waves what is a transverse wave
zaharov [31]

Answer:

A wave in which the medium vibrates at right angles to the direction of the propagation is called transverse wave.

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Two students make the following claims:
antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

F=m.g

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

W=F.s\cos\theta

where:

s= displacement of the object

\theta= angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

  • In case of free-fall the angle between the force is and the displacement is zero.
  • In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

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Electrons are found outside of the nucleus.
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