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frosja888 [35]
3 years ago
12

A roller coaster car starts at a height of 90 M at Point a and is moving at a speed of 20 m per second on a later Hill at Point

C assuming friction does not affect the motion of the car what is its height at Point C
Physics
2 answers:
ankoles [38]3 years ago
7 0
Hi there!

The answer you are looking for.....



20.4 m!





Hope this helps you!

Have a great day!
valentina_108 [34]3 years ago
3 0
The answer is 20.4m in height. You can use the kinematics equation Vf^2=V0^2+2ad  Therefore we have solving for d or height, we get d=Vf^2/2g
(20m/s)^2/(2*9.8m/s^2)=20.4m
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The density of mercury is 13.6 g/ml. what is its density in lbs/L
solong [7]

Answer:

can u send a picture to identify it

Explanation:

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4 0
3 years ago
PLEASE HELP ME!!!!! You draw an arrow from left to right on a piece of paper and place the paper well behind a glass of water. Y
julia-pushkina [17]

Answer:

I think it is the first one: <u>The arrow appears reversed because light is bent as it enters the water, and again as it exits. The two light paths cross, making the direction of the arrow appear crossed.</u>

Explanation:

7 0
2 years ago
Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The
Kay [80]

Answer:

25.33 rpm

Explanation:

M = 100 kg

m1 = 22 kg

m2 = 28 kg

m3 = 33 kg

r = 1.60 m

f = 20 rpm

Let the new angular speed in rpm is f'.

According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

Initial angular momentum = final angular momentum

(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

                                  (1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'

(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

2660 = 105 x f'

f' = 25.33 rpm

8 0
3 years ago
Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04%
Hoochie [10]

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]  

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N

3 0
3 years ago
Just like energy is <br> matter is lost through an ecosystem
Novosadov [1.4K]
Yes energy is reduced 
7 0
3 years ago
Read 2 more answers
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