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vodomira [7]
3 years ago
11

If you increase the frequency what happens to the distance between the waves (wavelength)?

Physics
1 answer:
Dahasolnce [82]3 years ago
5 0

Answer:

C)the distance decreases Shorter wavelength

Step by step Explanation:

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The product of 14 and a cubed
olya-2409 [2.1K]

Answer:

14 × a^3

Explanation:

Product means multiplication

Cubed means to the power of 3

6 0
4 years ago
Read 2 more answers
Calculate the time of fight for a horizontally launched projectile from a height of 20m above the ground with an initial velocit
Hoochie [10]

Answer:

0.5sec

Explanation:

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Method one

Using the second law of motion

H=ut-1/2gt^2

20=5t-1/2×10×t^2

20=5t-5t^2

dh/dt = 5-10t

where any constant is zero therefore the 20 is zero

5-10t=0

Collect like terms

-10t= -5

t=1/2 = 0.5sec

2nd method

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Using the time taken formula

t=u/g

t=5/10

t=0.5sec

4 0
3 years ago
Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He
Rufina [12.5K]
For the answer to the question above, 
 the distance from i to j is 5 parts 
(2 parts from i to k and 3 parts from k to j) 

The y distance from i to j is 
10 - 2 = 8 

Each part is 8/5 = 1.6 
Therefore the distance between the 2 parts from i to k is 3.2 

From the y coordinate of I which is 2 plus the 3.2 to point k 
2 + 3.2 = 5.2 

Answer y =5.2 

Now just convert that to fraction and that will be the answer
3 0
3 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
Radioactive decay of 40k atoms in an igneous rock has resulted in a radio of 25 percent 40k atoms to 75 percent 40Ar and 40Ca at
Ivanshal [37]
In order to answer this, we would need to know
the half-life of ⁴⁰K.  Perhaps when you look that up,
you'll be able to answer this on your own.

6 0
3 years ago
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