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2 years ago

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

Physics

1 answer:

Lesechka [4]2 years ago

3 0

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it. Because the contact between the bolt and the pliers working surface is less.

<h3>What is mechanical advantage ?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of the given mechanical machine.

The efficiency to open the stubborn bolt depends upon the contact between the working surface of the pliers and the bolt.

The contact between the bolt and the top pair of pliers working surface is less. Its mechanical advantage is less.

Hence, the top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

To learn more about the mechanical advantage, refer to the link;

brainly.com/question/7638820

#SPJ1

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An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

$\displaystyle v_o=\sqrt{2gh_m}$

$v_o=\sqrt{2\cdot 9.8\cdot 77.5}$

$v_o=39\ m/s$

(b)

The maximum time or the time taken by the object to reach its highest point is calculated as follows:

$\displaystyle t_m=\frac{v_o}{g}$

$\displaystyle t_m=\frac{39}{9.8}$

$t_m=4\ s$

7 0

3 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

kap26 [50]

Answer:

a) $\omega \approx 219.911\,\frac{rad}{s}$ , b) $\alpha = 16.916\,\frac{rad}{s^{2}}$ , c) $a_{t} = 1.776\,\frac{m}{s^{2}}$ , d) $a_{n} = 5077.889\,\frac{m}{s^{2}}$ , e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) $v = 23.091\,\frac{m}{s}$ .

Explanation:

a) The maximum angular velocity of the fan is:

$\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$\omega \approx 219.911\,\frac{rad}{s}$

b) The angular acceleration of the fan is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}$

$\alpha = 16.916\,\frac{rad}{s^{2}}$

c) The magnitude of the tangential aceleration is:

$a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)$

$a_{t} = 1.776\,\frac{m}{s^{2}}$

d) The magnitude of the centripetal acceleration is:

$a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)$

$a_{n} = 5077.889\,\frac{m}{s^{2}}$

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

$v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)$

$v = 23.091\,\frac{m}{s}$

5 0

3 years ago

A correlation coefficient represents what two things

kenny6666 [7]

Answer:

A correlation coefficient represents the following:

1- The direction of the relationship

2- The strength of the relationship

3 0

3 years ago

Calculate the number of coulombs per second if the area is 4cm, recombination rate of hole is 1000 cm/s and the differential len

musickatia [10]

Answer:

number of coulombs = 1.28 × $10^{-22}$

option b is correct

Explanation:

Given data

area A = 4cm

rate of hole r = 1000 cm/s

length L = 2mm

to find out

number of coulombs

solution

we will apply here number of coulombs formula that is

number of coulombs = e×A×L× r

put here value e = 1.6 × $10^{-19}$ coulombs

and A = 4 × $10^{-4}$ mand length = 2 × $10^{-3}$ m

number of coulombs = 1.6 × $10^{-19}$ × 4 × $10^{-4}$ × 2 × $10^{-3}$ × 1000

number of coulombs = 1.28 × $10^{-22}$

so option b is correct

3 0

3 years ago

What are some examples of helpful and unhelpful frictions?

ladessa [460]

Helpful friction would be the friction on something that you wanted to stop (a car) while unhelpful friction is friction on something that you want to keep going (a curling rock)

8 0

3 years ago