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2 years ago

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A 10.0 kilogram mass of iron on earth exerts a downward force of 98 Newtons. On the moon this same 10.0 kilograms of iron is wei

ghed and found to exert a downward force 16.3 Newtons. How is this possible ?

2 answers:

Len [333]2 years ago

6 0

Answer: D) The gravity on the moon is one sixth the gravity of Earth.

Explanation:

Anarel [89]2 years ago

3 0

Answer:

The decrease in the downward force that a mass of iron can exert on the moon versus the force it exerts on earth is due to:

<u>The force of gravity on Earth is greater than the force of gravity on the moon</u>.

Explanation:

To recognize the calculation within the statement, you must know that the Newton unit is equal to:

<u>Newton = (Kilogram * meter) / second ^ 2 </u>

And that the gravities of the Earth and the Moon are:

Earth gravity = 9.807 m / s ^ 2
Moon gravity = 1.63 m / s ^ 2

Finally, you must know the force formula (since we are talking about a descending force):

<u>Force = mass * acceleration</u> (gravity is a measure of acceleration)

Since the mass in both cases is the same (10 kilograms), the variation in acceleration will provide different values of descending force, as shown below, replacing the values:

Downward force on Earth = 10 Kg * 9,807 m / s ^ 2 = 98.07 Kg * m / s ^ 2 = <u>98.07 Newtons</u>.
Downward force on the Moon = 10 Kg * 1.63 m / s ^ 2 = 16.3 Kg * m / s ^ 2 = <u>16.3 Newtons</u>.

As you can see, <u>when it comes to force, the less acceleration (in this case less gravity), the lower the downward force will be with a mass of equal weight</u>.

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2 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

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Explanation:

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3 years ago

A 500-kilogram sports car accelerates uni-

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Explanation:

Given:

x₀ = 0 m

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2 years ago

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and

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Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

The height is h = 2.0 m

The angle is $\theta = 20^o$

The distance is $w = 10m$

The speed is $u = 11 m/s$

The coefficient of static friction is $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

$ma = mgsin \theta + F_f$

where $F_f$ is the frictional force mathematically represented as

$F_f =\mu F_x =\mu mgcos \theta$

where $F_x$ is the horizontal component of the force

substituting into the equation

$ma = mgsin \theta + \mu mg cos \theta$

$ma =mg (sin \theta + \mu cos \theta )$

making a the subject of the formula

$a = g(sin \theta = \mu cos \theta )$

substituting values

$a = 9.8 (sin(20) + (0.02 ) cos (20 ))$

$= 3.54 m/s^2$

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as

$sin \theta = \frac{h}{l}$

making $l$ the subject

$l = \frac{h}{sin \theta }$

substituting values

$l = \frac{2}{sin (20)}$

$l = 5.85m$

Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as

$v^2 =u^2 + 2 (-a)l$

The negative a means it is moving against gravity

substituting values

$v^2 = (11)^2 - 2(3.54) (5.85)$

$v= \sqrt{79.582}$

$= 8.92m/s$

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

Initial velocity along the x-axis which is mathematically evaluated as

$v_x = vcos 20^o$

substituting values

$v_x = 8.92 * cos (20)$

$= 8.38 m/s$

Initial velocity along the y-axis which is mathematically evaluated as

$v_y = vsin\theta$

substituting values

$v_y = 8.90 sin (20)$

$= 3.05 m/s$

Now the motion through the pool in the vertical direction can mathematically modeled as

$y = y_o + u_yt + \frac{1}{2} a_y t^2$

where $y_o$ is the initial height,

$u_y$ is the initial velocity in the y-axis

$a_y$ is the initial acceleration in the y axis with a constant value of ( $g = 9.8 m/s^2$ )

at the y= 0 which is when the height above ground is zero

Substituting values

$0 = 2 + (3.05)t - 0.5 (9.8)t^2$

The negative sign is because the acceleration is moving against the motion

$-(4.9)t^2 + (2.79)t + 2m = 0$

Solving using quadratic formula

$\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

$\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

$t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

$t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

$t = 1.02s$

At this time the position the motorcycle along the x-axis is mathematically evaluated as

$x = u_x t$

x $=8.38 *1.02$

$x =8.54m$

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool

7 0

3 years ago