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LuckyWell [14K]
2 years ago
6

What three sprites are needed to create the Space Invaders game that was discussed in the unit? spaceship, laser, enemy gnome, a

lien, star moon, sun, rocket spaceship, alien, black hole
Physics
1 answer:
Lilit [14]2 years ago
3 0

Answer: anlien, enemy gnome, spaceship

Explanation:

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Describe how reactivity changes as you go down Group 1A.
kenny6666 [7]

Answer:

it is a the answer is a btw

Explanation:

3 0
3 years ago
What is the force exerted by a solid surface that opposes gravity?
jonny [76]

Electromagnetic force between the molecules! 
6 0
3 years ago
Read 2 more answers
the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert
AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

P=\frac{F}{A} (1)

In this case we have the surface of a sole of a tennis shoe:

A=0.0290 m^{2} (2)

And the mass m of the person who wears it:

m=65 kg

On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

F=m.g=(65 kg)(9.8m/^{2})

F=637N (3)

Substituting (2) and (3) in (1):

P=\frac{637N}{0.0290 m^{2}} (4)

Finally:

P=21965.517 Pa This is the pressure the shoe exert on the ground

5 0
3 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
3 years ago
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0
3 years ago
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