Answer:
W = 0.49 N
τ = 0.4851 Nm
Force
Explanation:
The weight force can be found as:
W = mg
W = (0.05 kg)(9.8 m/s²)
<u>W = 0.49 N</u>
The torque about the pivot can be found as:
τ = W*d
where,
τ = torque
d = distance between weight and pivot = 99 cm = 0.99 m
Therefore,
τ = (0.49 N)(0.99 m)
<u>τ = 0.4851 Nm</u>
The pivot exerts a <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.
Ohm's Law tells the relationship between voltage, current, and resistance.
It can be written in three different ways, depending on which ones you know,
and which one you want to find.
Here's the one we need:
Resistance = (voltage) divided by (current)
= (120 V) / (0.5 Amp)
= 240 ohms .
According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^
<h3>What is collision and momentum?</h3>
The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).
<h3>Briefing:</h3>
From conservation of momentum
Initial momentum = final momentum
m u +M U =(m+M) V
2000×25 i^ +1500×30 j^ =(2000+1500) V
V = 8.735i^ + 11.25j^
Final velocity when they stick together is 8.735i^ + 11.25j^
To know more about Collide visit:
brainly.com/question/27993473
#SPJ4
The complete question is -
A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.
Explanation:
(a) For an isothermal process, work done is represented as follows.
W = 
Putting the given values into the above formula as follows.
W =
= 
= 
= 
= 29596.78 J
or, = 29.596 kJ (as 1 kJ = 1000 J)
Therefore, the required work is 29.596 kJ.
(b) For an adiabatic process, work done is as follows.
W = 
= 
= 
= 49.41 kJ
Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.
(c) We know that for an isothermal process,
or, 
= 
= 11 atm
Hence, the required pressure is 11 atm.
(d) For adiabatic process,
or, 
= 
= 28.7 atm
Therefore, required pressure is 28.7 atm.
