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Fittoniya [83]
2 years ago
12

How to find angular velocity from linear velocity and radius.

Physics
1 answer:
Lilit [14]2 years ago
6 0

Answer:

ω=v/r.

Explanation:

<em><u>angular velocity= linear velocity/radius</u></em>

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A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th
Alex Ar [27]

Answer:

 avriage force F = 2722.5 N

Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

      vf² = v₀² - 2 ax

The final carriage speed is zero (vf = 0)

      0 = v₀² - 2ax

      a = v₀² / 2x

      a = 1.1²/(2 0.200)

      a = 3.025 m / s²

      a = 3.0 m/s²

We calculate the average force

      F = ma

      F = 900 3,025

      F = 2722.5 N

3 0
3 years ago
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}

5 0
3 years ago
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

\Sigma \tau = 0

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
\tau = r \times F

Doing the summation using their respective lever arms:

0 = L Tsin\theta  - dF_g

dF_g = LTsin\theta

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o

Now, let's solve for 'T'.

T = \frac{dMg}{Lsin\theta}

Plugging in the values:
T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}

3 0
1 year ago
A moose runs through the woods and covers 825 m North in 118 s. What is the average velocity of the moose?
weqwewe [10]
  • Total displacement=825m
  • Total Time=118s

Average Velocity=Total Displacement/Total Time

\\ \sf\longmapsto Average\: Velocity=\dfrac{825}{118}

\\ \sf\longmapsto Average\:Velocity\approx 8m/s

4 0
3 years ago
It takes 1 minute for 45 c to pass a point in a circuit, what is the current flowing through the circuit?
ohaa [14]

Answer:

0.75 A

Explanation:

An electric current is a flow of charged particles.

A current is defined through its intensity, which is given by:

I=\frac{q}{t}

where

I is the current intensity

q is the charge passing through a given point in the circuit

t is the time it takes for the charge q to pass that point in the circuit

In this problem, we have:

q = 45 C is the charge passing through the point in the circuit

t=1 min = 60 s is the time elapsed

Therefore, the current intensity is:

I=\frac{45}{60}=0.75 A

5 0
3 years ago
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