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Deffense [45]
2 years ago
11

A boy throws a steel ball straight up. consider the motion of the ball only after it has left the boy's hand but before it touch

es the ground, and assume that forces exerted by the air are negligible. for these conditions, the force(s acting on the ball is (are
Physics
2 answers:
kap26 [50]2 years ago
3 0
The forces acting on the ball, aside from air friction, would be the force exerted on the ball by the boy when he threw it up, and gravity working against the motion of the ball
sweet-ann [11.9K]2 years ago
3 0
The gravity is working against the movement of the ball
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Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a
ArbitrLikvidat [17]

Answer:

4

Explanation:

The weight of the rock is W = mg = (80 kg) (10 m/s²) = 800 N.

The mechanical advantage is therefore 800 N / 200 N = 4.

5 0
3 years ago
Acceleration and Force
olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

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2 years ago
During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

7 0
3 years ago
Read 2 more answers
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
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QUIK!!!!!!!!!!!!Which is an activity that will build muscle endurance?
katen-ka-za [31]
There is many activities you can do like running, sit-ups, push-ups, and squats. 
7 0
3 years ago
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