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tekilochka [14]
3 years ago
13

Ann walks 80 meters on a straight line 33degrees north of east starting at point 1. Draw Ann's path. Represent Ann's walk with a

vector of length 80 meters.
Physics
1 answer:
zubka84 [21]3 years ago
7 0
The x-component of Anne's path:
80 x cos(33)
= 67.1 meters
y-component of Anne's path:
80 x sin(33)
= 43.6 meters
To draw he diagram, you draw a horizontal line starting at point 1 to the right representing 67.1 meters. At the end of this line, draw a vertical line upwards representing 43.6 meters.
Connect the start of the horizontal line at point 1 to the end of the vertical line with an angle of 33° between them.
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3 years ago
Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert
damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

F = k \frac{q\cdot q}{r^2}

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

q+\frac{q}{2}=\frac{3}{2}q

while the other charge will be

q-\frac{q}{2}=\frac{q}{2}

So, the new force will be

F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F

So, the force will decrease to 3/4 of its original value.

6 0
3 years ago
The electrons equal or copy the number of _____
Westkost [7]
Protons copy the number equal
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4 years ago
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What type of clouds are associated with low pressure?
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4 years ago
A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En
Maru [420]

Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M,  radius R and rotational inertia I  about its center of mass, rolling without  slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline,  we analyze this as follows:

The force of gravity (W = Mg) acting straight down  is resolved into components parallel and  perpendicular to the incline.

Since the object rolls without  slipping there is a force of  friction (Ff) acting on the object,  at it’s point of contact with the  incline, in the direction up  the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM   ⇒    m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction,    Ff = I*α / R

This is used in the expression  derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration  of the edge that contacts the incline by

a = R*α

Since the object rolls without  slipping this has the same  magnitude as aCM so we have  that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒  aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R²   (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0
3 years ago
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