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3 years ago

Objects in a composition occupy ____________.

Physics

2 answers:

Natasha_Volkova [10]3 years ago

7 0

Answer:

Positive space

Explanation:

Marat540 [252]3 years ago

4 0

<span>The answer here is positive space. Within a composition, any given object occupies positive space, whereas white space around this, or simply areas of blank space that are not being somehow used, is defined as negative space. Hopefully this clears up your question!</span>

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A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

Read 2 more answers

I need help with these Physics problems

adoni [48]

Answer:

1. 3 m

2. 27 s

Explanation:

1. "A car traveling at +33 m/s sees a red light and has to stop. If the driver can accelerate at -5.5 m/s², how far does it travel?"

Given:

v₀ = 33 m/s

v = 0 m/s

a = -5.5 m/s²

Unknown: Δx

To determine the equation you need, look for which variable you don't have and aren't solving for. In this case, we aren't given time and aren't solving for time. So look for an equation that doesn't have t in it.

Equation: v² = v₀² + 2aΔx

Substitute and solve:

(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx

Δx = 3 m

2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m. How long did it take to accelerate?"

Given:

v₀ = 0 m/s

a = 4.8 m/s²

Δx = 1800 m

Unknown: t

Equation: Δx = v₀ t + ½ a t²

Substitute and solve:

1800 m = (0 m/s) t + ½ (4.8 m/s²) t²

t ≈ 27 s

4 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

Three-Dimensional Thinking

gavmur [86]

Answer:

Explanation: both triple by 3

4 0

2 years ago

Kinetic energy has this unit?

Kobotan [32]

Heres your great lovley answer that you need

3 0

3 years ago