Answer:
a = 0.5195 m/s²
θ = 9.997º ≈ 10º
Explanation:
We apply Newton's 2nd Law as follows:
∑ Fx = m*ax
∑ Fy = m*ay
Then we have
∑ Fx = F₁x + F₂x = m*ax ⇒ 600*Cos 40º + 600*Cos (-20º) = 2000*ax
⇒ ax = 0.5117 m/s²
∑ Fy = F₁y + F₂y = m*ay ⇒ 600*Sin 40º + 600*Sin (-20º) = 2000*ay
⇒ ay = 0.0902 m/s²
the magnitude of the acceleration of the barge is
a = √(ax² + ay²) = √((0.5117 m/s²)² + (0.0902 m/s²))= 0.5196 m/s²
and the direction is
θ = Arctan (ay / ax) = Arctan (0.0902 / 0.5117) = 9.997º ≈ 10º
Answer:
C. Replacing one gas by another under the same conditions, has no effect on pressure.
Explanation:
Ideal gas:
A gas is treated as an ideal gas if temperature is high and pressure is low.
Kinetic energy for ideal gas given as
So when temperature of gas is increases then Average molecular kinetic energy will also increases.
The size of molecule is negligible as compare to the dimension of container. It mean that volume occupied by molecule is less as compare to the volume of container.
The between molecules is perfectly elastic.
Ideal gas equation
P V = m R T
So the option C is not always true.
Answer: Solids
Explanation: Solids have very little movement and the particles have barely enough movement to vibrate. We can't see it but the particles are indeed vibrating. Solids have the least amount of kinetic energy which is moving energy.
depending on the object the force pressing in on the object are equal with the quantity of the object.
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s
