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2 years ago

How speed and position affect the energy possess by an object?

Physics

1 answer:

Svetllana [295]2 years ago

4 0

Answer:

more speed means that an object has more energy, now if an object's place is something such as a hill, the potential energy will increase meaning an object will have more speed and acceleration. this is because you have the earth's gravity helping you out when the object goes downhill, giving it the higher potential energy

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Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor

Lemur [1.5K]

Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

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2 years ago

How do you appreciate sweating mechanism of human body to control the temperature of the body?

ehidna [41]

Explanation:

When the body temperature tends to rise, such as during physical exercise, the body begins to sweat. The sweat with high water content is secreted in the skin and when it evaporates into the environment, it cools the body. This is due to the property of water having high heat capacity. It carries with it a lot of heat per molecule (because water requires much energy – than most materials - for its temperature to rise by a degree) hence ideal for cooling. This is why on a hot day, sweating makes the skin feel cooler than the surrounding.

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3 years ago

A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car

Gekata [30.6K]

Answer:

1.63366

Explanation:

I got this answer from calculator soups physics calculators. I really recommend their website for formulas.

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2 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$