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viva [34]
4 years ago
15

How long does it take a cheetah that runs with a velocity of 34m/s to run 750m?

Physics
1 answer:
Eva8 [605]4 years ago
8 0

Answer:

It would take 5 seconds

Explanation:

I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.

V = d / t

34 = 170 / t

34 x t = 170

34t = 170

t = 170 / 34

t = 5

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A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate
PilotLPTM [1.2K]

Answer:

T(1)=21

Explanation:

The equation of the position in kinematics is given:

x(t)=x_{0}+v_{0}t+0.5at^{2}

  • x(0) is the initial position, in this it is 0
  • v(0) is the initial velocity (20 m/s)
  • a is the acceleration (2 m/s²)

So the equation will be:

x(t)=20t+0.5*2*t^{2}

x(t)=20t+t^{2}    

Now, the Taylor polynomial equation is:

f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

f(t)=x(t)=20t+t^{2}

f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t

f''(t)=\frac{dv(t)}{dt}=a(t)=2

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}

f(0)=x(0)=0

f'(0)=v(0)=20

f''(0)=a(0)=2

T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}

T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}

T(t)=20t+t^{2}

Let's put t=1 so find the how far the car moves in the next second:

T(1)=20*1+1^{2}

T(1)=21

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

4 0
3 years ago
and area' Explain the following in your answers, use the words pressure', 'force A truck used in the desert has wide tyres. 6 A
nlexa [21]

Answer:

a cricket stamp

Explanation:

4 0
3 years ago
How does gravity affect potential energy?
wel
This is more along the lines of "Does gravity affext potential energy" Sort of. Potential energy is an odd one to imagine, sometimes. It's the energy possessed by an object or system by dint of it's spatial and mechanical configuration. That definition alone is perhaps not so useful...and it's certainly not official. But what it means is that an object can potentially have energy due to where it is or what state the system is in. Imagine we have a box and it's on the floor. The box, for all intents and purposes, has no potential energy. It isn't going anywhere and it just sits on the floor. It can't do any work in it's current position. Now we hoist the box into the air. For any distance the box travels from the floor, it gains potential energy. Now let's back track. We've changed the box's spatial configuration by hoisting it into the air and so have given it potential energy. Why does it now have potential energy? Because we can now drop the box (costing us no energy) and the box will fall. Maybe it falls onto a passer-by and injures them. Box on the floor = No energy. We lift the box = We spend our energy and give the box potential energy (as it wants to fall toward the ground). We drop the box = Potential energy is converted to kinetic energy as the box falls. Box injures someone = The kinetic energy has done work upon the person. So we can see how it all flows and connects. We have to put energy into the box to fight against gravity, but you can't destroy or create energy....so the energy we've spent is potentially stored 'inside' the box. Clearly, gravity effects a LOT of potential energies around us. In fact to some small extent, it's probably impossible to entirely avoid it's effects.
5 0
3 years ago
. Which is the transfer of energy as waves moving through space?
Solnce55 [7]

Answer:

Radiation is the answer to your question

6 0
3 years ago
A 3 billiard ball (mass = 0.16 kg) moving at 4.0 m/s collides elastically head-on with a cue ball (mass = 0.17 kg) that is initi
alexandr1967 [171]

Answer: 1.9394m/s

Explanation: m1 (mass of 3 billiard ball) = 0.16kg

U1(initial velocity of the 3 billiard ball) = 4.0m/s

M2 (mass of the cue ball) = 0.17kg

U1 (initial velocity of the cue ball) = 0m/s

Final velocities of both the cue ball and 3 billiard ball after collision = ?

According to the principle of conservation of linear momentum, total momenta before collision = total momenta after collision:

M1U1 + M2U2 = (M1+M2)V

0.16*4.0 + 0.17* 0 = (0.16 + 0.17)V

0.64 + 0 = 0.33V

V = 0.64/0.33 = 1.9394m/s

8 0
3 years ago
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