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Gala2k
5 days ago
14

A car in a movie moving at 20 m/s flies off a cliff that is 74.3m tall. How long is the car in the air for and how far does the

car travel before it hits the ground below?
Physics
1 answer:
Softa [2.9K]5 days ago
4 0
We apply the formula S=½gt², where S represents the vertical distance of 74.3m, and g is the acceleration due to gravity valued at 9.8 m/s². Therefore, t=√(2S/g)=3.884 seconds. Consequently, the car remains airborne for around 3.894 seconds. Its horizontal journey is vt=20×3.894=77.88m.
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A model of a spring/mass system is 4x'' + e−0.1tx = 0. By inspection of the differential equation only, discuss the behavior of
Keith_Richards [3153]
Displacement stabilizes over time. It is known that exponentials raised to infinity approach zero, hence the system model will yield as time approaches infinity, resulting in 4x'' + e−0.1tx = 0. As time approaches infinity, we deduce that 4x'' equals zero. Consequently, upon integrating, we derive 4x' = c, and further integration leads to the conclusion 4x = cx + d.
4 0
8 days ago
A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it
Keith_Richards [3153]
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = \sqrt{rg} = \sqrt{2.80 \times 9.8} = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

P.E_{highest} = mgh

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as P.E_{highest} = mg(2r)

where u is the velocity at the lowest point. Consequently, the modified equation is

= \sqrt{v^{2} + 4gr}

= \sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = \frac{(m_{1} + m_{2})u}{m_{2}}

= \frac{(20 + 5) \times 11.71}{5}

= \frac{292.75}{5}

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.

3 0
8 days ago
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What
Ostrovityanka [3092]

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m


8 0
26 days ago
Read 2 more answers
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
inna [2995]
The answer is (C) 4 beats per second. The number of beats is computed as the difference between the frequencies of the two tuning forks. Plugging in the frequency values yields a result. Thus, the number of observable beats per second will be 4.
3 0
21 day ago
Lien is pushing a box across a table. She used a force of 100 N to get the box moving. Which force did she overcome to get the b
Yuliya22 [3237]
I will analyze each option. My assumption is that the answer is C.

Option A states that gravity acts downward on the box but does not affect its horizontal acceleration, provided there is no friction.

Option B indicates that the normal force goes upward on the box, which also does not influence horizontal acceleration.

In option C, the reaction force discussed relates to Newton’s 3rd law. This reaction force acts on Lien rather than the box itself, meaning she must overcome this force to set the box in motion. I believe this is the correct choice.

Option D refers to the push force applied by her; she wouldn’t have to counteract her own force regarding the box, but must address the reaction force as I mentioned in option C.
4 0
1 month ago
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