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4 years ago

Which leader maintained control of Cuba from 1934 to 1959?

Physics

1 answer:

Alexeev081 [22]4 years ago

8 0

The leader Fulgencio Batista maintained control of Cuba from 1934 to 1959.

Answer: B. Fulgencio Batista

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Question 3, need to know what is the minimum force to move the ball up (answer is 15N)

Lady_Fox [76]

The minimum amount of force needed to move the ball is 15N

8 0

3 years ago

Read 2 more answers

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 13 m/s, skate

kolbaska11 [484]

Let us say that:

1 = 1st player notation

2 = 2nd player notation (the opponent)

a. First let us establish the distance travelled by the 2nd player:

d2 = 13 m/s * (t + 1.5)

d2 = 13 t + 19.5

Then the distance of the 1st player:

d1 = v0 t + 0.5 a t^2 (v0 initial velocity = 0 since he started from rest)

d1 = 0.5 * 4 m/s^2 * t^2

d1 = 2 t^2

The two distances must be equal, d1 = d2:

2 t^2 = 13 t + 19.5

t^2 – 6.5 t = 9.75

Completing the square:

(t – 3.25)^2 = 9.75 + (- 3.25)^2

t – 3.25 = ±4.5

t = -1.25, 7.75

Since time cannot be negative, therefore:

t = 7.75 seconds

So he catches his opponent after 7.75 seconds.

b. Using the equation:

d1 = 2 t^2

d1 = 2 * (7.75)^2

d1 = 120.125 m

So he travelled about 120.125 meters when he catches up to his opponent.

3 0

3 years ago

The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed

padilas [110]

Answer:

Details in the explanation

Explanation:

Vertical Launch

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

$h = -16t^2 + 64t$

Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

$h = -16(1)^2 + 64(1)=64-16=48\ ft$

The object is at a height of 48 feet

At t=2 sec

$h = -16(2)^2 + 64(2)=128-64=64\ ft$

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

$h = -16(3)^2 + 64(3)=192-144=48\ ft$

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

$h = -16(4)^2 + 64(4)=256-256=0\ ft$

The object is at ground level and has returned to the launch point.

5 0

3 years ago

A ball is dropped from the top of a tall building h=5m about how long does it take for the ball to hit the ground.

pishuonlain [190]

D=-5m
a(gravity)=-9.8m/s^2
vi= 0m/s
t=?
use equation d=vi*t+0.5a*t^2
because vi=0, you can cross out vi*t because anything multiplied by 0= 0
rearrange the equation to say t^2=d/0.5a
t^2= -5/-4.9
t^2=1.02
find the square root...
final answer: t=1s

5 0

3 years ago

A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?

madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy $=\frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0

3 years ago