A plane has a mass of 2,000 kg and is traveling 100 meters above the ground. What is the plane’s potential energy?

Caitlin is riding her bike. Which of the following statements tells what is happening?

gtnhenbr [62]

What are the following statements? If there's one that mention a description of current action, or motion, that's your answer.

8 0

3 years ago

Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the

Andre45 [30]

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, $E=3\times 10^6\ V/m$

Let the electrical potential is, $V=10^8\ V$

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

$V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m$

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

5 0

3 years ago

The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d

d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of

SashulF [63]

Answer:

$0.2592 \ or \ 25.92\%$

Explanation:

The exponential density function is given as

$f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t$

$\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}$

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

$P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}| \ 0\leq t\leq 300$

$P(0\leq X\leq 300)=-0.7408+1\\=0.2592$

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

3 0

3 years ago