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4 years ago

How do scientists look for black holes in the sky

2 answers:

6 0

To find black holes in space, scientists use telescopes to study visible light, X-rays, and radio waves. When gas orbits a black hole, it gets very hot and starts emitting X-rays and radio waves. Through special telescopes, the area surrounding the black hole looks really bright.

pentagon [3]4 years ago

5 0

Scientists look through special telescopes and the area surrounding the black hole is suppose to look really bright.

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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

Pie

Answer:

a) v = 3,843 m / s, b) 46.7º North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East

initial instant. Before the crash

p₀ = mₐ vₐ₀

final insttne. After the crash

p_f = (mₐ + m_b) vₓ

p₀ = p_f

mₐ vₐ₀ = (mₐ + m_b) vₓ

vₓ = $\frac{m_a}{m_a + m_b} \ v_{ao}$

let's calculate

vₓ = $\frac{16.7}{16.7 + 29.3} \ 7.26$

vₓ = 2,636 m / s

direction north

initial p₀ = m_b v_{bo}

final p_f = (mₐ + m_b) v_y

p₀ = p_f

m_b v_{bo} = (mₐ + m_b) v_y

v_y = $\frac{m_b}{m_a+m_b} \ v_{bo}$

let's calculate

v_y = $\frac{29.3}{16.7 + 29.3} \ 4.39$

v_y = 2.796 m / s

the final speed of the two two vehicles is

v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{2.636^2 + 2.796^2}$

v = 3,843 m / s

b) let's use trigonometry to find the direction

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ (2,796 / 2,636)

θ = 46.7º

This direction is 46.7º North East

7 0

3 years ago

The amount of work required for a

kirill [66]

Answer:

if I'm not wrong, 0 Joules

Explanation:

If it is not being moved at all, then no work is being done

8 0

3 years ago

A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu

vredina [299]

Answer:

a) $v_1=5.5172\ m.s^{-1}$

b) $v_2=-1.5152\ m.s^{-1}$

c) $v_3=4.6154\ m.s^{-1}$

d) $v_{avg}=3.8462\ m.s^{-1}$

Explanation:

Given:

distance down the field in the first interval, $d_1=16\ m$

time duration of the first interval, $t_1=2.9\ s$

distance down the field in the second interval, $d_2=-2.5\ m$

time duration of the second interval, $t_2=1.65\ s$

distance down the field in the third interval, $d_3=24\ m$

time duration of the third interval, $t_3=5.2\ s$

velocity in the first interval:

$v_1=\frac{d_1}{t_1}$

$v_1=\frac{16}{2.9}$

$v_1=5.5172\ m.s^{-1}$

velocity in the second interval:

$v_2=\frac{d_2}{t_2}$

$v_2=\frac{-2.5}{1.65}$

$v_2=-1.5152\ m.s^{-1}$

velocity in the third interval:

$v_3=\frac{d_3}{t_3}$

$v_3=\frac{24}{5.2}$

$v_3=4.6154\ m.s^{-1}$

We know that the average velocity is given as the total displacement per unit time.

$v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}$

$v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}$

$v_{avg}=3.8462\ m.s^{-1}$

3 0

3 years ago

Prior Knowledge Questions (Do these BEFORE using the Gizmo.)A buoy is anchored to the ocean floor. A large wave approaches the b

dimaraw [331]

Answer:

bounce up and down

Explanation:

Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.

8 0

3 years ago

They occupy the space surrounding the nucleus of the atom, they have a -1 charge

kipiarov [429]

Answer:

Electrons.

Explanation:

Look it up.

4 0

3 years ago