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Galina-37 [17]
3 years ago
6

A 2.26 cm tall object is placed in 18.6 cm in front of a convex lens. The focal length

Physics
1 answer:
Gnesinka [82]3 years ago
7 0

Answer:

the image is -38.7 cm from the lens

Explanation:

We use the lens equation which says  

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}  $

where u is the distance to the object, v is the distance to the image, and f  is the focal length.

Now, in our case

u =18.6cm,

v= 35.8cm,

f = 35.8cm;

therefore,

$\frac{1}{18.6}+\frac{1}{v}=\frac{1}{35.8}  $

$\frac{1}{v}=\frac{1}{35.8}-\frac{1}{18.6}  $

\boxed{v=-38.7cm}

where the negative sign indicates that the image is virtual and on the same side of the lens as the object.

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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

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two cars go through 2 different crashes. car one 1 experiences a 500N impulse for a duration for 15s, while car2 experiences the
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Explanation:

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression :

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