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Ronch [10]
3 years ago
9

A passenger railroad car has a total of 8 wheels. Springs on each wheel compress--slightly--when the car is loaded. Ratings for

the car give stiffness per wheel (the spring constant, treating the entire spring assembly as a single spring) as 2.8×10^7N/m. When 30 passengers, each with average mass of 80 kg, board the car, how much does the car move down on its spring suspension?
Physics
1 answer:
deff fn [24]3 years ago
5 0

Answer:

Explanation:

mg = kx

x = mg/k

x = 30(80)(9.8)/2.8e7 = 0.00084 m  ≈ 1 mm

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How far can a person run in 15 seconds if she runs at an average speed of 160 cm/s? State your answer in meters.
uysha [10]

Answer: distance = 24 meters

Explanation:

speed = distance / time

distance = speed x time

speed = 160 cm/s = 1.6 m /s     [ 100 cm = 1 m, 160 cm = 1.6 m ]

time = 15 s

distance = 1.6 m/s x 15 s = 24 m

8 0
3 years ago
38 Points + Brainlyest!!!
Damm [24]
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8 0
3 years ago
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A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is its acceleration?
saul85 [17]
Alright let's start this off with our basic equation!

Accerlation (a) = ?

Initial velocity (V1)= 26 m/s
Final velocity (V2) = 0 ft/s because the car comes to a complete stop!

Time (t) = 6 seconds

The equation for acceleration is below!
a = \frac{Final velocity - Initial Velocity}{time}

So now, just plug in the values! 
a = \frac{0 - 26}{6}
a = \frac{-26}{6}
a = -4.33 m/s²

Therefore, your acceleration is -4.33 m/s²!! Hope this helped and was one of the branliest answers :') 

5 0
3 years ago
Is this object showing acceleration for the first 2 seconds? explain your answer.
iVinArrow [24]
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3 0
3 years ago
We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength
san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

\theta=1.22\frac{\lambda}{b}

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm

hence, the smallest diameter objective lens is 13.4cm

8 0
3 years ago
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