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lilavasa [31]
2 years ago
14

Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

can throw a small rock. one albert is 88 meters. how many square alberts is equal to one acre? (1 acre = 43,560 ft2 = 4050 m2)
Physics
1 answer:
True [87]2 years ago
7 0

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Therefore 1 acre is equivalent to about 0.52 square alberts.</span>

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A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
2 years ago
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
Aleksandr [31]
Let
upthrust = T
weight = W = mg
Air resistance = F

When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)

Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
3 0
2 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
2 years ago
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond
luda_lava [24]

Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

With 3 nodes       λ₃ = 4L / 5

The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

         f = v / λ

         f = v  n / 4L

Let's consider each length independently

L = 0.75 m

        f₁ = 340 1/4 0.75 = 113.33 n

         f₁ = 113.33 Hz

        f₃ = 113.33   3

       f₃ = 340 Hz

L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

       f₃ = 56.67 3

       f₃ = 170 Hz

8 0
2 years ago
A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw
navik [9.2K]

Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7  x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0
2 years ago
Read 2 more answers
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