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Afina-wow [57]
3 years ago
10

The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the gr

ound. After the exhibition, the watermelon is allowed to slide along to the ground along a smooth ramp. How high above the ground is the watermelon at the moment its kinetic energy is 4.61 kJ?
Physics
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

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1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

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2 years ago
Which of the following best defines insulator?
Leya [2.2K]

Answer:a substance with low ability or no ability to conduct energy

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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

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