What is done when an object is moved by a force

Parallax error occurs when the observer records data when he/she is at an angle to the event he/she is observing. Where do you t

kompoz [17]

Answer:Orient your line of sight directly above the measurement markings.

Explanation: parallax error is a type of systematic error that occurs when an observer views a measurement marking at a wrong angle. This causes a noticeable disparity in results obtained. Therefore the best way to prevent this error is to view and record the data from the correct angle. This can be obtained by:

-Place the measurement device on its edge so it is level with the object being measured.

-Seek out the finest possible edge of the measurement device, or use a device with finer edges.

In conclusion, ask other observers to also take the reading and get an average of their results. It can help cancel out parallax error results.

4 0

3 years ago

The wavelength of the light is 0.63 micrometers. How much of this length stays in 1 centimeter

bazaltina [42]

11,066,669.
hope it help.

7 0

3 years ago

A negatively charged particle is moving to the right, directly above a wire have a current flowing to the right. In which direct

Varvara68 [4.7K]

Answer:

C) upward

Explanation:

The problem can be solved by using the right-hand rule.

First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).

Now we can apply the right hand rule to the charged particle:

- index finger: velocity of the particle, to the right

- middle finger: direction of the magnetic field, out of the page

- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward

Therefore, the direction of the magnetic force is upward.

3 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .