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atroni [7]
2 years ago
7

Two identical stones are thrown from the top of a tall building. Stone 1 is thrown vertically downward with an initial speed v,

and stone 2 is thrown vertically upward with the same initial speed. Neglecting air resistance, how do their speeds compare on hitting the ground?
Physics
1 answer:
Molodets [167]2 years ago
5 0

If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

v= \frac{1}{2}gt

Where

g = Gravitational acceleration

t = time

As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.

Both bodies will end with the same thermal speed.

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1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
Which value is represented by the slope of the line?
netineya [11]

The slope of the line is

(change in ' y ' between the ends)  /  (change in ' x ' between the ends)

Slope = (630g - 0) / (70 cm^3 - 0)

Slope = (630 / 70) g/cm^3

<em>Slope =  9.0 g/cm^3</em>

5 0
3 years ago
The real advantage to hydrostatic weighing is that it __________.
katovenus [111]

The real advantage to hydrostatic weighing is that it gives one of the most accurate measurements of body fat.

7 0
3 years ago
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kogti [31]
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5 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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