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const2013 [10]
3 years ago
6

If you jump off a 7 m diving platform, how long will it take you to hit the

Physics
1 answer:
Allushta [10]3 years ago
3 0

Answer:

C.) T = 1.20s

Explanation:

From equation of motion,

S = ut + ½gt²

S = 7m

g = 9.8m/s²

U = 0m/s (since he starts from rest)

t = ?

S = ut + ½gt²

7 = 0 * t + ½ * 9.8 * t²

7 = 0 + 4.9t²

t² = 7 / 4.9

t² = 1.42857

t = √(1.42857)

t = 1.195s

t = 1.20s

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Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w
amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

1)

  • Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
  • There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

        v = \lambda * f (1)

  • Replacing by the givens, and solving for f, we get:

       f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)

⇒     f = 8.6 Ghz (with two significative figures)

2)

  • Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

  • Solving for t, we get:

       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

       ⇒  t = 0.2 ms (with two significative figures)

6 0
2 years ago
I need homework help
KIM [24]
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2

<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
</span>
5 0
3 years ago
Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.
babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

d\sin \theta =m\lambda

For Destructive interference

d\sin \theta =(m+\frac{1}{2})\lambda

where \lambda =wavelength

d=slit\ width

m=order of maxima and minima

for second order maxima i.e. m=2

For smallest separation taking \lambda =400 nm, \theta =90^{\circ}

d\sin 90=2\times 400\times 10^{-9}

d=0.8\times 10^{-6}

d=0.8\mu m

   

6 0
3 years ago
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga
Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0
3 years ago
If the distance between two masses is tripled, the gravitational force between changes by a factor of
maw [93]

A. 1/9

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}

6 0
3 years ago
Read 2 more answers
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