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2 years ago

Two 5.0 kilogram objects have different gravitational potential energies. What property of the objects must differ?

Physics

1 answer:

ipn [44]2 years ago

7 0

Answer:

Height.

Explanation:

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

$P.E = mgh$

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Hence, the property of the object (having a mass of 5 kilograms) which must differ to have different gravitational potential energies is the height from which they are falling from.

The object having the higher height would have a greater gravitational potential energy than the lower object.

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Mamont248 [21]

Answer:

1. 571.43m/s

2. 142.9m and 342.9m

Explanation:

1.Take the difference in time.

1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

0.6×571.43=342.9m

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2 years ago

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at

Aleks04 [339]

Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

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3 years ago

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Paul [167]

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It is important to take into account a system's constants, which cannot be altered by experiments or observations.

In conclusion, the mass of the water can be taken into consideration as a continuous system parameter.

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1 year ago

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snow_lady [41]

Answer:

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3 years ago

A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the strin

forsale [732]

Answer:

$\frac{m}{r}(v_b^2-v_t^2)+2mg$

Explanation:

When the ball is at the bottom position of the vertical circle, the forces acting on the ball are:

- The tension in the string, $T_b$ , upward

- The weight of the ball, $mg$ , downward

The resultant of these forces must be equal to the centripetal force, which points upward as well (towards the center of the circle), so:

$T_b-mg=m\frac{v_b^2}{r}$ (1)

where $v_b$ is the speed of the ball at the bottom of the circle, r the radius of the circle, and m the mass of the ball.

When the ball is at the top position of the vertical circle, the weight still acts downward, however the tension in the string also acts downward, so the equation of the forces becomes:

$T_t+mg=m\frac{v_t^2}{r}$ (2)

where

$T_t$ is the tension in the string in the top position

$v_t$ is the speed of the ball in the top position

By subtracting eq.(2) from eq.(1), we find:

$T_b-mg-(T_t+mg)=m\frac{v_b^2}{r}-m\frac{v_t^2}{r}\\T_b-T_t=\frac{m}{r}(v_b^2-v_t^2)+2mg$

So, this is the difference in tension between the two positions.

7 0

3 years ago

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