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Doss [256]
3 years ago
5

An object is initially moving at an unknown velocity. It accelerates at a rate of 1.5m/s2 to a new velocity of 60 m/s in 25 s. W

hat was the initial velocity?
Physics
1 answer:
Alexandra [31]3 years ago
6 0
<h3>Answer:  22.5 m/s</h3>

=====================================================

Work Shown:

acceleration = ( finalVelocity - initialVelocity )/(change in time)

1.5 = (60 - x)/(25)

1.5*25 = 60-x

37.5 = 60-x

x = 60-37.5

x = 22.5

The initial velocity is 22.5 m/s

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A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength
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Describe a life cycle of a star
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Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t
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According to newton's law of gravitation, the gravitational force(F)  is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg =  G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg =  G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

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5 0
1 year ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

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ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

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5 0
3 years ago
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