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OLEGan [10]
3 years ago
14

Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of

mass. a complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. such an approach can greatly simplify problem solving.
Physics
1 answer:
Nataliya [291]3 years ago
4 0
Treating the system as a point-like particle allows us to assign a quantity to the object and monitor this quantity throughout any changes. The complexity of the system which includes geometry, appearance, and extensions can complicate the studying of the system.
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Explain why the extrapolated temperature is used to determine the maximun temperature of the mixture rather than the highest rec
Oksi-84 [34.3K]
The extrapolated temperature is used to define the maximum temperature of the mixture relatively than the highest recorded temperature in which the conclusion will effect in a higher specific heat value. Heat is bound to escape from whatever apparatus is using, therefore it is needed to account for the loss of the heat that does not go into increasing the temperature of the mixture. 
6 0
3 years ago
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A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
Alla [95]

Answer:

v_{ox}= 19.6\ m/s

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = v_{ox}

Initial vertical velocity= v_{oy}  (Since ball was thrown horizontally only)

Acceleration acting horizontally, a_x = 0 m/s²  [ Since no acceleration acts horizontally) ]

Vertical Acceleration, a_y = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

H= v_{oy}t+\frac{1}{2}a_yt^2

5= (0)t+\frac{1}{2}(9.8)t^2

t= \frac{10}{9.8}=1.02 s

Then using Equations of motion for horizontal motion,

R= v_{ox}t+\frac{1}{2}a_xt^2

20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2

v_{ox}= 19.6\ m/s

4 0
3 years ago
A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the ball
ololo11 [35]
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.

Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³

Given:
At ground level,
p₁ = 752 mm Hg
     = (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
     = 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
     = 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
     = 300.8 K

At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
     = 9.7326 x 10³ Pa
T₂ = 235 K

If the volume at  36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
     = (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
     = 762.15 m³

Answer: 762.2 m³  
3 0
3 years ago
A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its
likoan [24]

Answer:

When X = 0

Speed = maximum    V (max) = ω A

Acceleration = zero       a(max) = - ω^2 A

From x = A sin ω t        sin = 0     so displacement = zero

V = ω A cos ω t             cos = 1 and speed = maximum

a = - A ω^2 sin ω t            sin = 0 and acceleration = zero

4 0
2 years ago
Why is the answer A and not C?<br>please help!​
MariettaO [177]

Answer:

because it gives a complete thought

4 0
2 years ago
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