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3 years ago

What is the acceleration at the apex of a vertical up and down problem?

Physics

1 answer:

Natalija [7]3 years ago

7 0

Answer:

The correct option is +9.81 m/s²

Explanation:

The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².

At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).

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A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

3 0

3 years ago

This whole worksheet. I am not good at this and I am completely lost

Ket [755]

1.) potential energy
2.)potential and kinetic
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3 0

3 years ago

A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a

sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

$v = \sqrt{\frac{T}{\mu} }$

where;

v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

$v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s$

Therefore, the speed of the sound wave on the string is 545.78 m/s.

3 0

2 years ago

According to Newton, which of these will determine if and how an object will move?

mr Goodwill [35]

Answer:

Explanation:

5 0

3 years ago

Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.

jasenka [17]

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

<h3>Force per unit length</h3>

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d - I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L = (μI/2π) x (I/2d + I/d)

F₃/L = (μI/2π) x (3I/2d)

F₃/L = 3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

Learn more about magnetic force here: brainly.com/question/13277365

#SPJ1

6 0

2 years ago