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adelina 88 [10]

3 years ago

5

A .2.-kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back and gives it a velocity of 14 m/s in

the opposite direction.

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

25(6+14) = 5

25(6+14) = 55/.02 = 500/2 = 250 N

25(6+14) = 55/.02 = 500/2 = 250 Na = F/m = 250 N/.25 = 1000 m/s^2

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Calculate the AMA of an access ramp if it takes 255 N of force to push a person in a wheelchair having a combined weight of 764

Andreyy89

Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N

Substitute the values in to the expression,
AMA = 764 / 255
AMA = 2.99

After rounding-off to the nearest tenth value, it would be 3

Finally, option C would be your answer.

Hope this helps!

7 0

3 years ago

If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha

vovikov84 [41]

If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.

6 0

3 years ago

Read 2 more answers

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

3 years ago

The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length

Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 390nm = 390*10^{-9}$

Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

$y_{r2} = 1.49\mu m$

Therefor, the airy disk radius ranges from $1.49\mu m$ to $11.4\mu m$

7 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago