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3 years ago

5

Genevieve was working on a science lab. In her lab, she was combining two chemicals and timing how long the reaction took to for

m a precipitate. She repeated the investigation a total of five times. Her times are shown below. trial time (min:sec) 1 08:37 2 07:59 3 08:16 4 08:21 5 08:03 The next day, the instructor told Genevieve that the standard time for her investigation should have been 8 minutes 7 seconds. Which of the following choices describes Genevieve's work in the lab?

1 answer:

dexar [7]3 years ago

5 0

Answer: Her test trials had a high level of accuracy but a low level of precision.

Explanation:

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While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room an

Bogdan [553]

Answer:

the time needed for her to close the door is 1.36 s.

Explanation:

given information:

Force, F = 220 N

width, r = 1.40 m

weight, W = 790 N

height, h = 3.00 m

angle, θ = 90° = π/2

to find the times needed to close the door we can use the following equation

θ = ω₀t + 1/2 αt²

where

θ = angle

ω = angular velocity

α = angular acceleration

t = time

in this case, the angular velocity is zero. thus,

θ = 1/2 αt²

now, we can find the angular speed by using the torque formula

τ = I α

where

τ = torque

I = Inertia

we know that

τ = F r

and

I = 1/3 mr²

so,

τ = I α

F r = 1/3 mr² α

α = 3 F/mr

= 3 F/(w/g)r

= 3 (220)/(790/9.8) 1.4

= 5.85 rad/s²

θ = 1/2 αt²

π/2 = 1/2 5.85 t²

t = 1.36 s

5 0

2 years ago

What is the maximum number of unpaired d electrons that an atom or ion can possess?

Sonbull [250]

Answer:

5

Explanation:

The d subshell has 5 orbitals, each capable of holding a maximum of two electrons. Hund's rule tells us that every orbital in a sub-level must first be singly occupied by electrons before any orbital is doubly occupied. Therefore five electrons will fill the five orbitals within the d subshell.

3 0

3 years ago

Please need help with this

PtichkaEL [24]

Through is the answer to your question

3 0

3 years ago

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A sinusoidal wave travels along a stretched string. A particle on the string has a maximum speed of 2.0 m/s and a maximum accele

JulsSmile [24]

Answer:

The amplitude of the wave is 0.02 m.

Explanation:

Given that,

Maximum speed = 2.0 m/s

Maximum acceleration = 200 m/s²

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{a_{max}}{v_{max}}$

Put the value into the formula

$\omega=\dfrac{200}{2.0}$

$\omega=100\ rad/s$

We need to calculate the amplitude of the wave

Using formula of velocity

$v_{max}=A\omega$

$A=\dfrac{v_{max}}{\omega}$

Put the value into the formula

$A=\dfrac{2.0}{100}$

$A=0.02\ m$

Hence, The amplitude of the wave is 0.02 m.

8 0

3 years ago

An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e

Alex_Xolod [135]

Answer:

a) $\omega = 10.407\,\frac{rad}{s}$ , b) $m = 4.617\,kg$ , c) $A = 1.355\,m$

Explanation:

a) The system have a simple armonic motion, whose position function is:

$x(t) = A\cdot \cos (\omega\cdot t + \phi)$

The velocity function is determined by deriving the position function in terms of time:

$v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)$

The acceleration function is found by deriving again:

$a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)$

Let assume that $t = 0\,s$ . The following nonlinear system is built:

$A\cdot \cos \phi = 0.660\,m$

$-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}$

$-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}$

System can be reduced by divinding the second and third expressions by the first expression:

$\omega \cdot \tan \phi = 18.636\,\frac{1}{s}$

$\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}$

Now, the last expression is divided by the first one:

$\omega = 10.407\,\frac{rad}{s}$

b) The mass of the block is:

$m = \frac{k}{\omega^{2}}$

$m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }$

$m = 4.617\,kg$

c) The phase angle is:

$\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega} \right)$

$\phi \approx 0.338\pi$

The amplitude is:

$A = \frac{0.660\,m}{\cos 0.338\pi}$

$A = 1.355\,m$

8 0

3 years ago

Read 2 more answers