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vivado [14]
3 years ago
15

A train travels 63 kilometers in 2 hours, and then 56 kilometers in 3 hours. What is its average speed? i need the answer and un

its please
Physics
1 answer:
kolbaska11 [484]3 years ago
4 0
The average speed is the ratio between the total space and the total time of the motion:
v= \frac{S_{tot}}{t_{tot}}
The total space is 
S_{tot}=63 km + 56 km = 119 km
while the total time is 
t_{tot}=2h+3h = 5h
So, the average velocity is
v= \frac{119 km}{5 h} =23.8 km/h


We can also rewrite it in m/s. The total space is S_{tot}=119000 m, while the time is t_{tot}=5h\cdot 3600 s/h = 18000s, and so
v= \frac{119000m}{18000s}=6.6 m/s
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What is the kinetic energy of a 50-kg child running to catch the school bus at
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A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

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