Answer: 70.5 km/h
Justification:
The question is not clearly stated but it seems you are asking for the x - component of the velocity of the helicopter.
You can find the x and y - components of the velocity using the trigonometric ratios sine and cosine.
The sine ratio relates the y-component and the velocity by:
sin(angle) = y-component of velocity / velocity
The cosine ratio related the x-component and the velocity by:
cos(angle) = x-component of velocity / velocity.
Since you have the angle and the velocity and are asked by the x-component of the velocity, you need to use the cosine ratio:
cos(35°)= x-component / 86.0 km/h
=> x -component = 86.0 km/h * cos(35°) = 70.5 km/h
Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:
Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same
Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:
As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:
b) If the temperatures changes, we can still reuse the ideal gas equation above:
Answer:
T = g μ_s ( M+m )
78.4 N
Explanation:
When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .
Net force on the lower box
= T - mg μ_s = M a ( a is the acceleration created by net force in M )
Considering force on the upper box
mg μ_s = ma
a = g μ_s
Put this value of a in the equation above
T - m gμ_s = M g μ_s
T = mg μ_s + M g μ_s
= g μ_s ( M+m )
2 )
Largest tension required
T = 9.8 x .50 x ( 10+6 )
= 78.4 N
<u>Answer:</u>
<em>Newtons II law: </em>
<em> </em>It is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>
Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.
Whenever the hanging weight moves downwards, the cart will accelerate to right side.
<em>For the hanging weight/mass</em>
When hanging weight of mass is m₁ and accelerate due to gravitational force g.
Therefore we can write F = m₁ .g
and the tension acts in upward direction T (negetive)
Now, Fnet = m₁ .g - T
= m₁.a
So From Newtons II law<em> F = m.a</em>
