Answer:
W has the lowest density and Y has the greatest density
Explanation:
Density of W = mass/volume = 11/24 = 0.45
Density of X = mass/volume = 11/12 = 0.91
Density of Y = m/v = 5.5/4 = 1.375
Density of Z = m/v = 5.5/11 = 0.5
From these we can find the answer......
Hope this answer is useful......
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y = 
t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m
★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
★ The speed of the hound and the hare
★ The speed of the hound and the hare = 25:18
As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
Now the distance travelled by the hare in it's 6 leaps..!
Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
The net force acting on the airplane is 25N.
Forces acting on the paper airplane when it is in the air:
- The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
- Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
- Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
- Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.
Given.
Weight of the paper airplane, F1 = 16N
The force of air resistance, F2 = 9N
Net force = F1 + F2
Net force = 25N
Thus, the net force acting on the airplane is 25N.
Learn more about the net force here:
brainly.com/question/18109210
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