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just olya [345]
3 years ago
12

How does kinect energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at 6

0 mph?
Physics
1 answer:
mote1985 [20]3 years ago
8 0

If it is the same vehicle, then the 60mph vehicle has more kinetic energy since it is moving faster. Therefore, it requires more energy to stop, and if it is the same car with the same beak system, the braking distance of the 30mph car will be significantly shorter than the 60mph car.



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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
4 years ago
Helpon
statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0
3 years ago
A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for
Elena-2011 [213]

Answer:

1) 10.1 s  2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

  • In the vertical direction, as the bomb is dropped, its initial velocity is 0.
  • So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

       \Delta y = \frac{1}{2}*a*t^{2} (1)

  • where Δy = -500 m (taking the upward direction as positive).
  • a=-g=-9.8 m/s²
  • Replacing these values in (1), and solving for t, we have:

       t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s

  • The time required for the bomb to reach the earth is 10.1 s.

2)

  • In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
  • As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

       x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.

3)

  • The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

  • In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

        v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s

5)

  • If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
  • So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0
4 years ago
Un bloque de 750 kg es empujado hacia arriba por una pista inclinada 15º respecto de la horizontal. El coeficiente de rozamiento
kotegsom [21]

Answer:

4776.98 N is the minimum force to start the rise.  

Explanation:

We can use the first Newton's law to find the minimum force to move the block.

So we will have:

F-W_{x}-F_{f}=0

Where:

  • F is the force
  • W(x) is the weight of the block in the x direction, W = mg*sin(15)
  • F(f) is the static friction force (F(f) = μN), μ is the static friction coefficient 0.4.

F=W_{x}+F_{f}=mgsin(15)+\mu N

F=mgsin(15)+\mu mgcos(15)

F=mg(sin(15)+\mu cos(15))

F=750*9.81(sin(15)+0.4*cos(15))

F=4776.98 N

Therefore 4776.98 N is the minimum force to move the block.

I hope it helps you!

8 0
3 years ago
Two friends, Joe and Sam, go to the gym together to strength train. They decide to start off with an exercise called flat bench
astra-53 [7]

Answer:

216.31 (the work done by gravity is -216.31) positive for going up.

Explanation:

We look at this question first by getting the right equation for <em>work</em>.

Which should be... W = F x D.

From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.

What is the formula for the force of gravity on an object near the earth?

Right here --- F_{grav} = mg

m = the mass and...

g  =  the acceleration due to gravity which is <em>9.81 m/s2</em>

Before we plug things in though, we need to convert everything to SI units,

the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.

Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) =  216.31 J or 216.31 N m.

3 0
3 years ago
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