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sergejj [24]
2 years ago
12

A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t

he biker take to travel this segment?
Physics
1 answer:
lawyer [7]2 years ago
7 0

Answer: 0.016 h

Explanation:

\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

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An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
kodGreya [7K]

Find the given attachment for solution

7 0
1 year ago
A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i
aniked [119]

Answer:

a. N = 2.49W b.  0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ  = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

6 0
1 year ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
2 years ago
A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.
Degger [83]
Hi, thank you for posting your question here at Brainly.

To compute for the change in potential energy, the equation would be:

delta PE =  mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta Pe = 0.98 J

The potential energy is converted to kinetic energy.
3 0
2 years ago
Read 2 more answers
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

   potential energy store has been transferred into its kinetic energy

   store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final speed, v_{0} in the initial speed and m

is the mass

→ v = 20 m/s and v_{0} = 0 m/s

→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0
1 year ago
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