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Nataly [62]
3 years ago
8

Explain why a roller coaster’s second hill cannot be taller than the first hill.

Physics
2 answers:
koban [17]3 years ago
8 0

The reason behind the second hill in the roller coaster not taller than the first hill is the frictional force acting on the roller coaster.

<u>Explanation: </u>

The height of the first hill in the roller coaster is taller with the decrease in the subsequent hills as the frictional forces reduces the kinetic energy.  

Coming down the first tallest hill, the speed and kinetic energy both increases as the motion is in line with gravity, but due to conservation of energy, the potential energy decreases. And as the potential energy decreases, the height of hill has to be decreased in order to complete the motion.

max2010maxim [7]3 years ago
4 0
There will not be enough momentum from the first hill to cross another hill if he same or larger size because of the way potential energy and kinetic energy works it will not be able go as high as it could go on he fist hill.
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Mr. Patel is looking for his cat in the garden at night. It is completely dark outside. Mr Patel shines his torch at the tree. W
UNO [17]

Answer:

the tree's light reflects into his eyes.

I think this is the ans

Explanation:

3 0
2 years ago
Using the graph predict how many paper clipsa 7.5 v battery would pick up for both the 25 coil electromagnet and the 50 coil ele
enyata [817]

The graph between the strength of the magnet(number of paper clips picked) and battery is approximately a straight line.

For 25 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 5 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

3 0
3 years ago
If two cars A and B are moving with velocity 60 km/hr and 80 km/hr
Vitek1552 [10]

Answer:

VAB = 20km/hr

Explanation:

<u>Given the following data;</u>

Velocity of car A, VA = 60km/hr

Velocity of car B, VB = 80km/hr

To find the relative velocity of B w.r.t A, VAB;

Since the two cars are moving in the same direction, we have;

VAB = VB - VA

Substituting into the equation, we have;

VAB = 80 - 60

<em>VAB = 20km/hr</em>

Therefore, the relative velocity of car B with respect to car A is 20 kilometers per hour.

3 0
3 years ago
A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of
Sholpan [36]
F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
a= \frac{F}{m}
a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
4 0
3 years ago
Read 2 more answers
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

4 0
3 years ago
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