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Marina86 [1]
3 years ago
6

What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita

tional potential energy at the top of the track? Assume a frictionless surface.
Physics
1 answer:
sesenic [268]3 years ago
4 0

Answer:

60m/s

Explanation:

initial energy = final energy

g.p.e = k.e

k.e = 0.5 × mass × velocity²

g.p.e = 990000J as per Question

990000Nm = 0.5 × 550 × V²

V² = 3600

V = 60m/s

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If an object's kinetic energy is zero, what is its momentum?
hammer [34]
If the object's kinetic energy is zero, then due to in multiplication factor, it's momentum will also be equal to zero 'cause the velocity of the object must be Nil

In short, Your Answer would be: "Zero"

Hope this helps!
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3 years ago
If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be
natima [27]
The equilibrium constant will be lowered and the equilibrium will shift to the left if the heat being produced is not removed.
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Read 2 more answers
An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , & c
Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

f = \frac{v}{2L}

inserting the values

f = \frac{343}{2(1.6)}

f = \frac{343}{2(1.6)}

f = 107.1875 Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

3 0
3 years ago
When you stand outside your are standing in the, Stratosphere. Troposphere. Mesosphere no
AVprozaik [17]
Uhhh I’m not really sure of the answer i think it’s stratosphere
8 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
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