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NemiM [27]
3 years ago
10

A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it

exert on the track
Physics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

46620\ \text{N}

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Normal force at the point will be

N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}

The force exerted on the track is 46620\ \text{N}.

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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole​
Aleksandr-060686 [28]

Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

4 0
1 year ago
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
3 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

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