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myrzilka [38]
2 years ago
14

What is the average speed of a car that traveled 300 meters in 5.5 seconds?

Physics
1 answer:
allochka39001 [22]2 years ago
4 0

Answer: 54.55 m/s

Explanation:

Distance = 300 m

Time = 5.5 s

Average Speed = Total Distance/Total Time

= 300/5.5

= <u>54.55 m/s</u>

Hope this helped..

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What is the mechanical advantage of the wheel and axle shown below?
goldenfox [79]

Answer:

B. 120

Explanation:

24/.2 = 120

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3 years ago
El conductor de un coche de 650 kg que va a 90 km/h frena y reduce su
Ivanshal [37]
A. 90km/h = 25 m/s
W=1/2mv^2
=1/2* 650* 25^2
= 203125(J)
b. v’= 90-50= 40km/h
= 100/9 m/s
W=1/2mv^2
=1/2*650*(100/9)^2
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7 0
3 years ago
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla
dusya [7]

Answer:

C). U_f = \frac{U_0}{2}

Explanation:

As we know that capacitance of a given capacitor is

C = \frac{\epsilon_0 A}{d}

now we know that energy stored in the capacitor plates

U_0 = \frac{Q^2}{2C}

here if all the dimensions of the capacitor plate is doubled

then in that case

C' = \frac{\epsilon_0 (4A)}{2d}

here area becomes 4 times on doubling the radius and the distance between the plates also doubles

So new capacitance is now

C' = 2C

so capacitance is doubled

now the final energy stored between the plates of capacitor is given as

U_f = \frac{Q^2}{2C'}

so the final energy is

U_f = \frac{Q^2}{4C}

U_f = \frac{U_0}{2}

4 0
3 years ago
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a
sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

v = \sqrt{\frac{T}{\mu} }

where;

v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s

Therefore, the speed of the sound wave on the string is 545.78 m/s.

3 0
2 years ago
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