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3 years ago

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A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

of 3.0 m. What is its angular momentum if its radius is 7.4 cm? (I know the answer is 9.6 kgm^2/s but I don't understand how to get it)

1 answer:

stira [4]3 years ago

4 0

Answer:

$9.5 kg m^2/s$

Explanation:

The angular momentum of an object is given by:

$L=mvr$

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

$L=(0.35)(9.0)(3.0)=9.5 kg m^2/s$

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Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

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Answer:

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When the roller coaster going down

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D. plasma ..................

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Read 2 more answers

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

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