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vodka [1.7K]
3 years ago
11

A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird f

eeder. How does the added mass affect the period of oscillation of the bird feeder?
Physics
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

The added mass will mean a longer period of oscillation.

Explanation:

The period of oscillation here is given by the formula;

T = 2π√(m/k)

Where m is mass and k is spring constant

From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.

Thus, the added mass will mean a longer period of oscillation.

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Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
marta [7]
First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:

For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²

For constant velocity:
d = constant velocity*time

The solutions is as follows:

   a = v,final - v,initial /t
  3.8 = (v₁ - 0)/4.6 s
  v₁ = 17.48 m/s

    Total distance = d1 + d2 + d3
    d1 = d = v,initial*t + 1/2*at²
    d2 = constant velocity*time
    
   Total distance =  0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
   d3 = 56.69 m

3 0
3 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
A wire with a resistance of 20 is connected to a 12 battery. What is the current flowing through the wire?
Dima020 [189]
15.49 should be the answer if that is 12 watt battery.
3 0
3 years ago
At which part of its orbit is the planet traveling at the slowest speed?<br> A, B, C, or D
Natali [406]
I think its D im not sure tho
3 0
3 years ago
Read 2 more answers
Determine the elastic energy U stored in<br> the compressed spring.
hoa [83]

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

7 0
2 years ago
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