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2 years ago

Give examples of not useful high friction

1 answer:

6 0

Friction can be bad by being too strong or too weak.

Sometimes, when it is too strong, it decreases efficiency since some energy is wasted and turns to heat. Friction can also damage equipment/objects like when you slide it on the floor.

When friction is too weak, like for instance when there is black ice- our center of gravity is displaced too quickly and we can fall. Likewise, if there is a lot of slush on the ground, cars can slip and slide.

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If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l

kakasveta [241]

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

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2 years ago

What are the base units in the metric system?

Delicious77 [7]

meter, millimeters, kilometers. liters. kilograms. centimeters etc... look up the rest

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3 years ago

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An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds.

dimaraw [331]

it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m

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2 years ago

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Does a paddle boat’s wheel rotate clockwise or counterclockwise? PLS HELP

finlep [7]

Answer:

clockwise and counter clockwise

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2 years ago

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

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2 years ago