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Tom [10]
3 years ago
14

A mass suspended from a spring is oscillating up and down as indicated. Consider the following possibilities. A At some point du

ring the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative); B At some point during the oscillation the mass has zero velocity and zero acceleration; C At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration; D At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative). Which possibility or possibilities occur?
Physics
1 answer:
dangina [55]3 years ago
7 0

A mass suspended from a spring is oscillating up and down, (as stated but not indicated).

A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative).  <em>Yes. </em> This statement is true at the top and bottom ends of the motion.

B). At some point during the oscillation the mass has zero velocity and zero acceleration.  No.  If the mass is bouncing, this is never true.  It only happens if the mass is hanging motionless on the spring.

C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration.  <em>Yes.</em>  This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass.  At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .  

D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative).  No.  This can only happen if the mass is hanging lifeless from the spring.  If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.  

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PLEASE HELP ASAP
morpeh [17]
A is the correct answer
7 0
3 years ago
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
3 years ago
2. A hanging wind-chime on a calm day would have kinetic or potential energy?
rjkz [21]

Answer:

it would have potential energy

7 0
3 years ago
A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
A toolbox of mass 3.2kg is lowered by a rope from the roof to the ground. Find the acceleration of the toolbox when the force of
Lisa [10]

Answer:

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

                              a + g = F / m

                              a = F/m - g

a)                            a = 40/3.2 - 9.81

                              a = 2.69 ≈ 2.7 m/s²   positive up

b)                            a = 20/ 3.2 - 9.81

                              a = 6.25 - 9.81

                                  = - 3.56 ≈ - 3.6 m/s²  negative down

8 0
3 years ago
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