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blsea [12.9K]
3 years ago
11

A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break

s as the elevator accelerates. What was the magnitude of the elevator's minimum acceleration? and What was the direction of the elevator's minimum acceleration?
Physics
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

2.4 m/s^2

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

T=m(g+a)

Where g=9.8 m/s^2

Substitute the values

156=12.8(a+9.8)

9.8+a=\frac{156}{12.8}=12.2

a=12.2-9.8=2.4 m/s^2

Hence, the acceleration =a=2.4 m/s^2

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

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i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

          F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

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            i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

           λ = m / L

           m = λ L

we substitute

           i B = λ g

           i = \frac{ \lambda \ g}{B}

let's calculate

          i = 0.2 9.8 / 0.4

          i = 4.9 A

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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
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Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

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Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

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