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3 years ago

10

You and your dog are playing tug-of-war. For a while, neither of you are winning but after a few hard tugs, your dog pulls the t

oy from your hands. What happened in this scenario? *

1 answer:

777dan777 [17]3 years ago

3 0

The dog used a greater force then me

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3 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 6.7 rev/

Alexxandr [17]

Answer: 7.095 rev

Explanation: since the motion of the washer is of a constant angular acceleration, then Newton's laws of motion applies.

Recall that

ω = ω0 + αt

Where α = angular acceleration = ?

ω = final angular velocity = 6.7rev/s

ω0 = initial angular velocity = 0 rev/s

t = time taken =8.8s

Since the motion of the dryer starts from rest, ω0 = 0, we have that

ω = αt

6.7 = α(8.8)

α = 6.7/8.8

α = 0.76 rev/s²

This value α = 0.76 rev/s² is a constant value and it is the same all through the motion.

When the lid of the washer was lifted, the motion was of a constant deceleration.

Recall that

θ = ω0t + αt²/2

Where θ = angular displacement

Since the washer is now decelerating, the final velocity (ω= 6.7 rev/s) of the accelerating motion now because the initial velocity of the decelerating motion.

Hence ω0 = 6.7 rev/s, α = - 0.76 rev/s² ( negative sign to denote deceleration) and t = 16.5s

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2 years ago

Read 2 more answers

When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh

aev [14]

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

$v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

$0-(V)^2=2\times (-g)\cdot H$

$H=\frac{V^2}{2g}$

i.e. maximum height is dependent on square of initial velocity

for twice the height

$2H=\frac{(V')^2}{2g}$

on comparing

$V'=\sqrt{2}V$

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3 years ago