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Ber [7]
3 years ago
10

How can free runners extend the time of impact when they land on the ground after jumping long distance

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

By leaping.

Explanation:

When free runners run, they also leap. They do this so they can get more distance and air time. Leaping will allow them to get more distance per jump.

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A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge
Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance C_1 with energy of 7.54 J, is connected in parallel with another capacitor C_2 , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

U_O=q_0^2/2C_1=7.54 J\\

The energy stored in the electric field is the sum of the energies of the two capacitors:

U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2

since the charge equally distributed,  q_1 = q_2 = q_o/2. and since they are connected in parallel the potential difference on both of them is the same :

V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\

hence,

U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J

8 0
3 years ago
When the weight of the object increase block what is the force of friction applied? Explanation?
erik [133]

Answer:

There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.

8 0
2 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
As the average kinetic energy of a substance increases, the tempature of the sample
taurus [48]
I have no clue. Sorry bro
6 0
3 years ago
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A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the
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I had the same question

6 0
3 years ago
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