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Pani-rosa [81]
3 years ago
5

Assuming that the hill on the left will provide all of the potential energy for the ride, would this roller coaster actually mak

e it around the loop?
Thanks!

Physics
1 answer:
trapecia [35]3 years ago
3 0
Yes I believe it would I think the answer is D
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Which of the following statements are TRUE about nuclear energy?
san4es73 [151]

Answer:

d

Explanation:

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3 years ago
An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in
SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

h=-16t^2+v_ot

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

-16t^2+250t=500

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

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How far will a car travel in 30 min at 40 m/s? (kilometres)
kondor19780726 [428]

Answer:

72km

Explanation:

30 mins --> 30 x 60 s = 1800 s

Distance --> Speed x Time

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6 0
3 years ago
A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0
3 years ago
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