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erica [24]
3 years ago
5

An automobile starts from rest and travels down a straight section of road. The distance s (in feet) of the car from the startin

g position after t seconds is given by s(t)=5t3 (a) Find the instantaneous velocity (in feet per second) at t=4 seconds.
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer: 240 ft/sec

Explanation:

If the distance s, of the car regarding his starting position, is defined as a function of time, by definition, the instantaneous velocity at any time, is given by the derivative of the displacement (distance from origin) respect the time, as follows;

v = ds / dt ⇒ v= d/dt (5 t³) = 15 t² = 15 t²

Replacing by t=4 s, we get:

v = 15. 4² = 15.16 = 240 ft/sec

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kipiarov [429]

Answer:

Uranus and Neptun are outer planets od the Solar system, since they are located after the asteroid belt. All of these outer planets are much larger then the inner ones so they are called the "ice giants". The other reason for this name is that they are very far from the Sun, so their temperature is low. Another feature they have in common is their atmosphere which is composed of gases, including methane, which is responsible for their blue color, since methane absorbs red light. However Neptune is known for very fast winds and storms in its atmosphere which is responsible for its high activity and changes.

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5 0
3 years ago
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and
spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

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