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3 years ago

Pulsars "blink" because they _____.

1 answer:

4 0

A pulsar, or a pulsing star, is a highly magnetized neutron star that emits a beam of electromagnetic radiation. So they blink when they are rotating because the beam of radiation they emit can only be seen when it is facing the Earth.
Hope this helps.

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Một chất điểm đang chuyển động với vận tốc 10(m/s) thì tăng tốc sau khi đi được 20(s) thì vật có vận tốc 20 (m/S)

Makovka662 [10]

Answer:8huhyuj

uhuhjuhnjknjnm

Explanation:

nuhyuihtyi8uij

5 0

2 years ago

Melting butter is a physical change. Which best describes what is happening?

kati45 [8]

Answer:

Option B

Explanation:

The butter forms a new arrangement of atoms i.e from solid to liquid

3 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree

Mnenie [13.5K]

The y-component of the acceleration is $0.22 m/s^2$

Explanation:

The y-component of the acceleration is given by

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction $\theta=50.0^{\circ}$

v = 4.65 m/s is the final velocity, in a direction $120^{\circ}$

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

$u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s$

So the y-component of the acceleration is

$a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

2 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

2 years ago

A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike th

alexira [117]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2. So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

4 0

3 years ago