50 POINTS‼️‼️‼️‼️‼️

Why do astronomers find it difficult to detect individual exoplanets?

Maurinko [17]

Light from the stars, because the orbits make it difficult to see them.

6 0

2 years ago

Which describes the greenhouse effect? 1. Earth continues to get warmer over time because it is slowly moving closer to the sun.

Alekssandra [29.7K]

Answer: The green house effect is best described by option 4 (Energy given off by earth is reflected off of earth's atmosphere back down to the surface).

Explanation:

The green house effect can be described as the energy given off by earth is reflected off of earth's atmosphere back down to the surface.

When energy from the sun passes through the atmosphere, some are absorbed which keeps the earth surface warm. While the rest is reflected back largely by cloud.

The energy which is emitted from the earth surface is called the infrared radiation. Some of the infrared radiation passess through the atmosphere but most is absorbed and re- emitted in all directions by the greenhouse gas molecules and clouds. This effect warms the earth surface and the lower atmosphere. Therefore this statement (Energy given off by earth is reflected off of earth's atmosphere back down to the surface) is correct about greenhouse effect.

For the greenhouse effect to occur, greenhouse gas molecules are mostly needed. Examples of these gases include:

--> Carbon dioxide (CO2),

--> Water vapor (H2O), and

--> Methane (CH4)

Over the years, the excessive human activities has lead to increase in the greenhouse gas molecules which has negatively affected the greenhouse effects.

6 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

a circuit has a total resistance of 5 ohms and a current of 0.3 amperes. What voltage must the battery supply?

iren [92.7K]

Answer:

1.5volts

Explanation:

V=IR =0.3×5=1.5volts

5 0

3 years ago

What happens to the wavelength of a wave if you increase the frequency?

dezoksy [38]

The product of (wavelength) times (frequency) is always the same number ...
the speed of the wave in whatever material it's traveling through. So if the
frequency is increased, then the wavelength must <em><u>de</u></em>crease by the same
factor, in order to keep the product the same.

4 0

3 years ago