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3 years ago

A car moves with an average speed of 45 miles/hour. How long does the car take to travel 90 miles?

Physics

2 answers:

Sloan [31]2 years ago

7 0

Answer:

Explanation:

If you are going 45 miles per hour, you will travel 45 miles in 1 hour. It will take you 2 hours to travel 90 miles at a speed of 45 miles per hour.

to make it easier multiply 45 by 2 = 90

45 x 2 = 90

Elis [28]3 years ago

5 0

Answer:

Explanation:

45+45=90

45 per hour

so, 2 hours

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How do you do this question? Please include free body diagrams and clear explanation, so I can understand.

vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0

3 years ago

Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?

Sliva [168]

Answer:

$\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star$ ...

$\huge\boxed{\fcolorbox{white}{blue}{mgh=4000}}$

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7 0

3 years ago

To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

7 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

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Which of the following changes would make a heat engine waste more energy as heat

Komok [63]

A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.

4 0

3 years ago