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lesantik [10]
4 years ago
12

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

rmine the minimum height h of the hill crest so that the car travels around both inside the loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take ???????? = 7.5 m and ???????? = 5 m.
Physics
1 answer:
sweet [91]4 years ago
7 0

Answer:

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

F_n = 1.72 \times 10^4

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

v = \sqrt{5 R g}

also by energy conservation this is gained by initial potential energy

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

so we will have

\sqrt{2gh} = \sqrt{5Rg}

now we have

h = \frac{5R}{2}

here we have

R = 7.5 m

so we have

h = \frac{5(7.5)}{2}

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

now when it reach point C then the speed will be

mgh - mg(2R_c) = \frac{1}{2]mv_c^2

v_c^2 = 2g(h - 2R_c)

v_c = 13.1 m/s

now normal force at point C is given as

F_n = \frac{mv_c^2}{R_c} - mg

F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)

F_n = 1.72 \times 10^4

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docker41 [41]

Answer:

While the image this question refers to is missing, I am assuming it shows the word FIRE (1) written in a normally readable way, (2) written in a mirrored way. The display (2) is used in order for other vehicle drivers being able to read the word FIRE in their mirrors (rearview and side).  The word printed in a mirrored way will appear normal via a mirror (i.e., the mirror "undoes" the mirroring originally introduced).

4 0
3 years ago
A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?
kap26 [50]

Answer:

26621 km

Explanation:

We are given;

Mass: m = 5.98 x 10^(24) kg

Period; T = 43200 s

Formula for The velocity(v) of the satellite is:

v = 2πR/T

Where R is the radius

Formula for centripetal acceleration is;

a_c = v²/R

Thus; a_c = (2πR/T)²/R = 4π²R/T²

Formula for gravitational acceleration is:

a_g = Gm/R²

Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²

Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;

Centripetal acceleration = gravitational acceleration.

Thus;

4π²R/T² = Gm/R²

Making R the subject gives;

R = ∛(GmT²/4π²)

Plugging in the relevant values;

R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))

R = 26.621 × 10^(6) m

Converting to km, we have;

R = 26621 km

7 0
3 years ago
How do scientists identify how old rock and fossils are?
slavikrds [6]

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3 years ago
A car is traveling 80km/h is 130m behind a truck traveling at 75km/h. How long will it take the car to reach the truck?
rjkz [21]
The car approaches the truck at a speed of 80-75=5km/h.  To travel the 130m, it takes
\frac{0.130km}{5 \frac{km}{h} } =0.026h=93.6s
8 0
3 years ago
1. How much heat energy is required to raise the temperature of a 5 kg aluminium bar
Ray Of Light [21]

Answer:

180 kJ

Explanation:

Given that:

Mass (m) = 5 kg

Initial temperature (T1) = 28°C

Final temperature (T2) = 68°C

The change in temperature (ΔT) = T2 - T1 = 68°C - 28°C = 40°C

Specific heat capacity of aluminium (c) = 900 J/kg°C

The quantity of heat energy required (q) is given by:

q = mcΔT

q = 5 kg × 900 J/kg°C × 40°C

q =  180000 Joules

q = 180 kJ

Therefore 180 kJ is required to raise the temperature of aluminium from 28°C to 68°C.

5 0
3 years ago
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