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tankabanditka [31]

1 year ago

10

A leopard of mass 65 kg climbs 7 m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g = 10 N/k

g.

1 answer:

Kay [80]1 year ago

3 0

Mass of Leopard = 65 kg

Height = 7 m

P.E = ?

We know,

P.E = mgh

= 65 * 10 * 7=4550 J

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calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

1 year ago

Read 2 more answers

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence = 38.0°

For yellow light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}$

${sin\theta_2}=0.9377$

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}$

${sin\theta_2}=0.9395$

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

${sin\theta_{critical}}=\frac {n_r}{n_i}$

Where,

${\theta_{critical}}$ is the critical angle

$n_r$ is the refractive index of the refractive medium.

$n_i$ is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

${sin\theta_{critical}}=\frac {1}{1.523}$

The critical angle is = sin⁻¹ 0.6566 = 41.04°

5 0

2 years ago

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet

Mumz [18]

<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb $\sqrt{b^2 + L^2}$ ) -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{25.0004}$ )

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

5 0

2 years ago

A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0

1 year ago

Read 2 more answers

Nora tied a string around a tennis ball, and then she swung it in a circle in front of her to demonstrate a planet orbiting the

777dan777 [17]

<em>Choice-A </em>is a true statement. The string holding the ball in orbit is a contact force, whereas gravity is a non-contact force.

Choice-C is half-true. The string keeps the distance between the bodies constant, but gravity doesn't "MAKE" the distance vary.

8 0

1 year ago

Read 2 more answers