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What object best represents a true scale model of the shape of earth?

BabaBlast [244]

Any object that is spherical in shape would best represent a true scale model of the shape of the Earth. Examples are ping pong balls, billiard balls, marble and other smooth spherical objects. The shape of the Earth is called the oblate spheroid. The "oblate" would refer to an oblong shape and "spheroid" would refer to an almost spherical shape. The earth has on almost spherical shape and has a slightly oblong appearance. The diameter from the South pole to the north pole was measured to have a value of 12714 km while the diameter of the equator is approximately 12756 km. As you can see, the values are not equal. This makes the earth not a perfect sphere.

6 0

3 years ago

Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc

motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

$E = mc^2$

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

$E = mc^2$

4 0

3 years ago

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

pulling one of the balls away from the other. To pull a ball away, click on the ball and drag it. What happens to the force arro

Soloha48 [4]

Gravity.: Gravity is the force that acts at a right angle to the path of an orbiting object.

7 0

3 years ago

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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

5 0

3 years ago