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3 years ago

What is the potential energy of a 2 kg ball 15 m in the air?

Physics

1 answer:

oee [108]3 years ago

7 0

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

Hope this helps you

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g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and

cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

= 20(11.8)

= 236N

4 0

4 years ago

Is it true or false that air pollution only occurs as a result of human activity

gregori [183]

Hi alexander it is very true air pollution is only caused by human activity

8 0

3 years ago

Read 2 more answers

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

If a weight hanging on a string of length 5 feet swings through 5^\circ on either side of the vertical, how long is the arc thro

Zielflug [23.3K]

<span>First we can find the circumference of the whole circle with a radius of 5 feet. circumference = 2 pi radius circumference = (2 pi) (5 feet) circumference = (10 pi) feet From one high point to the other high point, the string moves through an angle of 10 degrees. Since a full circle is 360 degrees, this angle is 1/36 of a full circle. Therefore, the arc length is 1/36 of the whole circumference. arc length = (1/36) (circumference) arc length = (1/36) (10 pi) feet arc length = 0.873 feet</span>

7 0

3 years ago

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

Leya [2.2K]

Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)

4 0

3 years ago