All categories

3 years ago

What is a bright streak of light that results when a meteoroid burns up in earth's atmosphere?

1 answer:

4 0

<span>When a meteoroid passes through Earth's atmosphere it's traveling very fast. The friction of the air makes gets the surface so hot it begins to burn or glow red</span>

You might be interested in

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb

GuDViN [60]

Answer:

Q = 12540 J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

$Q=mc\Delta T$

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

$Q=50\times 4.18\times (60-0)\\Q=12540\ J$

So, 12540 J of energy is used to heat the water.

7 0

2 years ago

The conversion of sugar to energy in the presence of oxygen is

Ahat [919]

Answer:

A. respiration.

Explanation:

Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage.

Additionally, mitochondria provides all the energy required in the cell by transforming energy forms through series of chemical reactions; breaking down of glucose into Adenosine Triphosphate (ATP) used for providing energy for cellular activities in the body of living organisms.

Basically, oxygen goes into the body of a living organism such as plants, humans and animals when they breathe while glucose is absorbed by the body when they eat.

Hence, the conversion of sugar to energy in the presence of oxygen is respiration.

8 0

3 years ago

What changes in airplane longitudinal control must be made to maintain altitude while the airspeed is being decreased?

garik1379 [7]

Explanation:

The changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is

We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.

8 0

3 years ago

Anyone knows this? Please answer... Spam will be reported.

Yakvenalex [24]

Answer:

The correct option is;

The assertion is correct, but reason wrong

Explanation:

The question is with regards to the relationship between work, energy, power, and velocity

The mass of each of the persons running up the staircase = Different

The time it takes each person to run up the stairs = Equal time

Let, 'm₁' and 'm₂' represent the mass of each of the persons that ran up the stairs and m₁ > m₂

Let 't' represent the equal time it takes then to run up the stairs

Let 'h' represent the height of the stairs

The energy, 'E', it takes to run up the stairs is equal to the potential energy, P.E., obtained at the top of the stairs

P.E. = m·g·h

Where;

m = The mass of the person at an elevated height

g = The acceleration due to gravity = Constant

h = The height reached above ground level

Given that the height reached is the same for both of the persons, we have

For m₁, P.E.₁ = m₁·g·h and for m₂, P.E.₂ = m₂·g·h

Therefore, where, m₁ > m₂, we have;

P.E.₁ > P.E.₂

∴ E₁ > E₂

Power, 'P', is the rate at which energy is expended

∴ Power, P = E/t

∴ P₁ = E₁/t > P₂ = E₂/t

Therefore, the person with the greater mass, 'm₁', uses more power than the person of mass 'm₂', in running up the stairs

Therefore, the assertion is correct

The average velocity, vₐ = (Total distance traveled, d)/(Total time taken, t)

Given that the distance, 'd', covered in running up the stairs by both persons is the same, and the time it takes them to complete the distance, 't', is also the same, we have;

The average velocity of the person with the greater mass m₁ is the same as the average velocity of the person with mass, m₂

Therefore, the reason is wrong

The answer is that the assertion is correct, but reason wrong

6 0

2 years ago