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3 years ago

What is the acceleration at the apex of a vertical up and down problem?

Physics

1 answer:

Natalija [7]3 years ago

7 0

Answer:

The correct option is +9.81 m/s²

Explanation:

The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².

At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).

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What is newton's second law of motion?

lord [1]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

3 0

2 years ago

A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N

crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force = ?

F = Applied force = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

6 0

3 years ago

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
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [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

3 years ago

1. According to paragraph 3 in the text, MOST of the electromagnetic waves from

Fed [463]

Answer:

Most of the EM waves from the sun that reach Earth are infrared waves, visible light, and UV radiation.

Explanation:

I hope this helps! Have a good day!

5 0

2 years ago

Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001

Vinvika [58]

Answer:

520000 or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0

2 years ago