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4 years ago

What is the compare the strength of static,sliding,and rolling friction

1 answer:

7 0

The force needed to overcome sliding friction is more than the force needed to overcome rolling friction or static or even fluid

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Instrumento que se utiliza para medir fuerzas

Klio2033 [76]

Answer:

Un medidor de fuerza

A force gauge for all you English speakers

Explanation:

Un medidor de fuerza es un instrumento de medición que se utiliza para medir fuerzas.

A force gauge is a measuring instrument used to measure forces.

8 0

3 years ago

An object is thrown vertically and has a speed of 25 m/s when it reaches 1/4 of its maximum height above the ground (assume it s

Crazy boy [7]

Answer:

$v_{i} =28.86\frac{ft}{s}$

Explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

$(v_{f} )^{2} =(v_{i} )^{2} -2*g*y$

Where:

$v_{f}$ : final speed in ft/s

$v_{i}$ : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For $v_{f} = 25\frac{ft}{s}$ , $y=\frac{1}{4} h$ ; where h is the maximum height

for y=h, $v_{f} =0$

Problem development

We replace $v_{f} = 25\frac{ft}{s}$ , $y=\frac{1}{4} h (ft)$ in the formula (1),

[ $25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}$ Equation (1)

in maximum height(h): $v_{f} =0$ , Then we replace in formula (1):

$0=(v_{i} )^{2} - 2*g*h$

$2*g*h=(v_{i} )^{2}$

$h=\frac{(v_{i})^{2} }{2g}$ Equation(2)

We replace (h) of Equation(2) in the Equation (1) :

$25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2} }{2g} }{4}$

$25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2} }{4}$

$25^{2} =\frac{3}{4} (v_{i} )^{2}$

$v_{i} =\sqrt{\frac{25^{2}*4 }{3} }$

$v_{i} =28.86\frac{ft}{s}$

7 0

4 years ago

A small car with a mass of 800kg moving with a velocity of 27.8 m/s. The car stops at a yellow light in 3.9 seconds. What force

raketka [301]

Answer:

F = 5702.56 N

Explanation:

Given that,

Mass of a small car, m = 800 kg

Initial speed of the car, u = 27.8 m/s

Final speed, v = 0

Time, t = 3.9 s

We need to find the force did it take for the car to stop.

The force acting on an object is given by :

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{800\times (0-27.8)}{3.9}\\\\F=-5702.56\ N$

So, the magnitude of force acting on the car to stop is 5702.56 N.

4 0

3 years ago

A 38.5kg man is in an elevator accelerating downward. A normal force of 343n pushes up on him. what is his acceleration?

alexira [117]

Answer:

<h3>The answer is 8.91 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{343}{38.5} = \frac{98}{11} \\ = 8.909090...$

We have the final answer as

Hope this helps you

4 0

3 years ago

A particular atom becomes more energetically stable when it loses a single electron. It is most likely an atom of which of the f

Scorpion4ik [409]

Answer:

Magnesium (Mg)

Explanation:

because i checked

6 0

2 years ago