It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0

3 years ago

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An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of

Georgia [21]

Answer: Velocity terminal = 0.093m/s

Explanation:

1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)

= (0.0604/2 - 0.06/2)m

= 2×10^-4

Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L

= (π×0.06×0.4)m²

= 0.075m²

Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.

Shearing stress = u×V.terminal/h = 0.86×V/0.0002

= 4300Vterminal

Therefore, Fw = shearing stress × A

30N = 4300Vterminal × 0.075

V. terminal = 30/4300 m.s

V. terminal = 0.093m/s

4 0

3 years ago

An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

4 0

3 years ago

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0

3 years ago