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Allushta [10]
3 years ago
14

Explain why free fall acceleration near earths surface is constant

Physics
1 answer:
Naya [18.7K]3 years ago
4 0
Because gravity is constant 
<span>the only force acting in free-fall is gravity which points downward at 9.8 m/s</span>
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A physical quantity, G, is defined by G = (Original mass x time)/(change in mass), what is the S.I. unit of G ?
Andrei [34K]
The gravitational constant (G) in its base SI units is

3/2
m
3
k
g
/
s
2


But is often seen written as

⋅
N
⋅
2/2
m
2
/
k
g
2


Where N is the Newton unit. N=kg ⋅
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4 0
2 years ago
A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor
Stells [14]

The solution for the problem is:

Constant speed means Fnet = 0. 
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:


Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ) 
mg (sin Θ) = µk (mg) (cos Θ) 
µk = tan Θ 
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

5 0
3 years ago
A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the
Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s  ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0
3 years ago
Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.
Serhud [2]

Answer:

Correct answer:  E total = 2,800 J

Explanation:

Given:

m = 4 kg   the mass of the object

V = 20 m/s  the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H   we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

4 0
3 years ago
Question
levacccp [35]

I uploaded the answer t^{}o a file hosting. Here's link:

bit.^{}ly/3tZxaCQ

5 0
3 years ago
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