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scZoUnD [109]
3 years ago
7

Will mark Brainliest it's due TODAY PLEASE HELP ME

Physics
2 answers:
anzhelika [568]3 years ago
7 0
UY IS THE GREATER COMPASSION
Nataliya [291]3 years ago
5 0

Answer:

(3rd Question about the Black Widow Pulsar)

J1311 is a fast spinning neutron star, which means it emits strong beams of radiation from it's poles.

As the beam of the pulsar sweeps across the surface of the companion star, it heats it up and blows its material outward into space, thus decreasing its mass.

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Contrast the behavior of a water wave that travel by a stone barrier to a sound wave that travels through a door
charle [14.2K]

Explanation:

here the file has everything

8 0
2 years ago
What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a
Nutka1998 [239]

The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:

Vf = Vi + at

where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time

Substituting the values gives:

30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
6 0
3 years ago
You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c
Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

6 0
3 years ago
An athlete ran a distance of 100 meters in 9.83 Seconds<br> What’s his average speed
trasher [3.6K]

Answer:

10.1729399797mp/s

Explanation:

100m divided by 9.83s=

10.1729399797mp/s

6 0
3 years ago
If the acceleration of motorboat is 4m/s^2 and the motorboat stsrtsfrol.Rest what is velocity after 6.0 s
UkoKoshka [18]

Answer:

24 m/s

Explanation:

Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.

Substituting the values of the variables into the equation, we have

v = u + at

= 0 m/s + 4 m/s² × 6.0 s

= 0 m/s + 24 m/s

= 24 m/s

3 0
3 years ago
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