Ask question

Ask question

All categories

3 years ago

13

The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard. Which

correctly describes the results? All of the darts hit the center of the board. Some of the darts hit the center of the board. The darts hit the same general area of the board. The darts hit very different areas of the board.

2 answers:

hoa [83]3 years ago

8 0

Answer:

The darts hit very different areas of the board.

Explanation:

Pavel [41]3 years ago

3 0

Answer:

The darts hit the same general area of the board.

Explanation:

The meaning of precision is the closeness of measured values to other measured values. Therefore, since the results of the dart game were precise and inaccurate, the darts would hit the same general area because the measured values should be close to one another.

You might be interested in

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Scientists use this part of the electromagnetic spectrum to diagnose broken bones. X–rays infrared waves gamma rays ultraviolet

Snezhnost [94]

X-rays are used to identify broken bones

4 0

2 years ago

Read 2 more answers

When developing a question for a scientific inquiry, the question will ideally

Lisa [10]

When developing a question for a scientific inquiry, the question will ideally :

- be measurable
- Will have various variables that could be used for further testing
- not just simply ask why

hope this helps

7 0

2 years ago

Read 2 more answers

PLEASE HELPPPPPPP!!!!!!!!!!

natka813 [3]

Answer:

F' = 800 N

Explanation:

The electrical force between charges is 400 N.

The electrical force between two charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If q₁' = 2q₁, new force becomes,

$F'=k\dfrac{q_1'q_2'}{r^2}\\\\F'=k\dfrac{2q_1\times q_2}{r^2}\\\\F'=2\times F\\\\F'=2\times 400\\\\F'=800\ N$

So, the new force becomes 800 N.

8 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Alexandra [31]

Mechanical waves need material stuff to travel through.

They cannot travel through vacuum.

That's why we can't hear the explosions on the sun.

That's also why astronauts standing on the moon or working outside the Space Station have to use radio to talk to each other, even if they're only a few inches apart. There's no air between them, or any other material stuff. So sound can't travel between them.

7 0

3 years ago

Read 2 more answers