I can't seem to figure out the angle between T1 and T2. So suppose, it is 10º; then T2 makes an angle of 35º w/r/t horizontal, and T1 makes an angle of 45º.
Sum the moments about the base of the crane; Σ M = 0. 0 = T2*cos35*L*cos40 + T1*cos45*L*cos40 - T2*sin35*L*sin40 - T1*sin45*L*sin40 - W*(L/2)*sin40 - T1*L*sin40 → length L cancels where W = 18 kN
0 = 0.259*T2 - 43kN T2 = 166 kN
<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>
Answer:
C. nuclear fusion, because the equation shows two hydrogen nuclei combining to form a helium nucleus
Explanation:
Nuclear reaction can either be; fission or fusion. Nuclear fission is the process by which a massive nucleus breaks in to two smaller nuclei of almost the same size with the release of high amount of energy. Nuclear fusion is the process by which two nuclei reacts, joins, to produce a massive nucleus (compared to the masses of the reacting elements) with the release of high amount of energy.
From the given equation, two hydrogen isotopes; deuterium and tritium reacts with each other to produce helium nucleus and a neutron.
This reaction is a nuclear fusion which produces a massive nuclei.
Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
Transverse waves travel on a direction that is perpendicular to the motion of the particles (or whatever medium is waving) So the particles must be moving east to west, which is transverse to the north-south motion of the wave
