An eagle flies from its perch in a tree to the ground to capture and eat its prey. describe its energy transfromation

Marina86 [1]

Hey! If this is on big ideas get a app called “Slader” It has answers to all math problems just like up the book you use > save it > type in the section you are doing > and look for the problems you are doing.

8 0

3 years ago

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s

8 0

3 years ago

Help! I don’t really know what it’s asking

Misha Larkins [42]

You have to do the math of each and see which one adds up to 66.5

6 0

3 years ago

A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?

vovangra [49]

Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

b) what is the frequency of the third-longest wavelength = 75Hz

Explanation:

The steps and appropriate formula and substitution is as shown in the attached file.

5 0

3 years ago

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

Question 6:

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3>The density of the mixture:</h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

Question 7:

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

Question 8:

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

Question 9:

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

Question 10:

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

3 years ago