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2 years ago

What would changing the frequency of a wave do to the wave?

1 answer:

3 0

The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.

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A 4.00-kg model rocket is launched, expelling 64.0 g of burned fuel from its exhaust at a speed of 675 m/s. What is the velocity

olga_2 [115]

Answer:

The velocity of the rocket is 7.8 m/s

Explanation:

3 0

2 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

Plss s ss s s s s s s s s

Crazy boy [7]

Answer:

i font know ewan madaling sabihin yan

8 0

3 years ago

Read 2 more answers

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

What combines to form bonds in compounds?

Mila [183]

A chemical compound consists of 2 or more chemical elements. They can be separated into simpler substances. Atoms of elements can be combined to form compounds. Molecules are groups of 2 or more atoms held together by chemical bonds. Minerals are chemical compound that can be found in nature. Valence is a measure of the combining power of an element.Answer: A. Atoms of elements.

8 0

3 years ago