What is the first step in using a balance beam scale?

torisob [31]

B. The angle will cause the suns rays to be reflected downward to the other scout

4 0

3 years ago

A boy on the edge of a vertical cliff 20-m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the grou

Komok [63]

Answer:

Horizontal distance from the foot of the cliff is 40 m

Explanation:

Given :

Speed of stone $v = 20\frac{m}{s}$

Vertical distance $d = 20$ m

Gravitational acceleration $g = 10$ $\frac{m}{s^{2} }$

From the kinematics equation,

$d = v_{o} t + \frac{1}{2} at^{2}$

Where $v_{o} =$ initial velocity ( $v_{o} = 0$ )

So time take to fall down,

$t = \sqrt{\frac{2d}{a} }$

$t = \sqrt{4}$

$t = 2$ sec

Now find horizontal distance travel by stone,

$x = vt$

$x = 20 \times 2$

$x = 40$ m

Therefore, the horizontal distance travel by stone is 40 m

3 0

3 years ago

What is KBr (periodic table) def

adelina 88 [10]

Answer: Answered already

Explanation:

5 0

3 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago