There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
None of the choices is an appropriate response.
There's no such thing as the temperature of a molecule. Temperature and
pressure are both outside-world manifestations of the energy the molecules
have. But on the molecular level, what it is is the kinetic energy with which
they're all scurrying around.
When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture. The work done
during the compression pumps energy into the molecules, their kinetic
energy increases, and they begin scurrying around fast enough so that
when they collide, they're able to stick together, form a new molecule,
and release some of their kinetic energy in the form of heat.
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows

![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)



T = 81.52 Deg C
Force acting during collision is internal so momentum is conserve
so (initial momentum = final momentum) in both directions
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Let Vx is and Vy are final velocities of car in +x and +y direction respectively.
initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)
- 750*25 + 1150*0 = (750+1150)
Vx
initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)
750*0 - 1150*5 = (750+1150)
Vy
from here you can calculate Vx and Vy
so final velocity V is
<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is
<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)
</span><span>
kinetic energy loss in the collision = final KE - initial KE</span>