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elena-14-01-66 [18.8K]
3 years ago
14

During the first part of this lab, we want to determine how the object distance is related to what two quantities

Physics
1 answer:
Nataly [62]3 years ago
4 0

Answer and Explanation:

The computation of the object distance related to two quantities is shown below:

It could find out by using the lens formula which is shown below:

\frac{1}{v}  - \frac{1}{u} = \frac{1}{f}

where,

v = image distance

u = object distance

f = focal length

It could be found by applying the above formula i.e considering the image distance, object distance and the focal length

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Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/
user100 [1]

Answer:

answer below

Explanation:

Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v= 739/162

v= 4.56

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3 years ago
What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?
Alisiya [41]
We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
\Delta V = I R
where
\Delta V is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and R=330 \omega, so the potential difference is
\Delta V = IR=(4.00 A)(330 \omega)=1320 V
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4 years ago
A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles
egoroff_w [7]

60 mph east...........

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3 years ago
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What kind of the question is that. We aren't really buried. We either go to Heaven or the other place. Some people say it's over when your buried in the ground but believers don't really think that. 
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3 years ago
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