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makvit [3.9K]
3 years ago
15

A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h

Physics
1 answer:
g100num [7]3 years ago
8 0

Answer:

54 km/hr

Explanation:

m/s to km/hr => 18/5

15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

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A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___
kondaur [170]
<span>Actually newtons third law says for every action there is an equal and opposite reaction, Hence here in this case, the diver diving of a raft is the action, after which surely reaction should come in the form where the raft and the driver will rebound with same speed back, and hence here the action force is diving and reaction force is rebounding from the diving place, with same intensity.</span>
5 0
3 years ago
Read 2 more answers
A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t
nexus9112 [7]

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m

So, the hayride will cover a distance of 68.18 m.

6 0
3 years ago
Which of the following situations has more kinetic energy than potential energy?
san4es73 [151]

Answer: I think its C

Explanation:

5 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
1. How are each of Newton’s Laws of motion demonstrated
stepan [7]

Answer:

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Newton's law of acceleration -  "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."

Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."

Hope this helps! god bless :)

Please give me brainliest!

5 0
3 years ago
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