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RUDIKE [14]
3 years ago
13

How can a driver best be prepare to enter sharp curves

Physics
1 answer:
Eduardwww [97]3 years ago
7 0
Slow down before entering the curve. Stay close to the outside. Use the gas pedal as you enter the middle of the curve.
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A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P)
Virty [35]

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

6 0
3 years ago
WOULUJUTUL RECIPECUIUS.
3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

Thus, we get finally,

                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

As we know, nano denoted by letter 'n' equals to 10^{-9}

So the force acting between two objects is 19.73 nN.

7 0
2 years ago
What is the net force acting upon this object? *
AnnZ [28]

2 N Right

.. .. .. .. .. .. .. ..

4 0
3 years ago
Read 2 more answers
If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?
Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy mgh, and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature {\Delta}T then will be

{\Delta}T=\frac{{\Delta}Q}{mc} .

Where m is the mass of water in the pool, c is the specific heat capacity of water, and {\Delta}Q is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0
3 years ago
What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds?
slava [35]

Answer:

The tangential velocity of a rotating object is:

v = r*w

where r is the radius, and w is the angular velocity.

w = 2*pi*f

where f is the frequency.

We know that the record plater does 11 revolutions in 20 seconds, then it does:

11 rev/20s = 0.55 rev/s = f

then we have:

w = 2*pi*0.55 s^-1 = 2*3.14*0.55 s^-1 = 3.454 s^-1

The radius of a record player is really variable, it is around 10 inches, so i will use r = 10in, which is the rotating part of the record player.

then the tangential velocity is:

v = 10in*3.454 s^-1 = 34.54 in/s

7 0
3 years ago
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