The tension in the rope B is determined as 10.9 N.
<h3>
Vertical angle of cable B</h3>
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
<h3>Angle between B and C</h3>
θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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Answer: 8.33m/s²
Explanation:
Mass of the airplane pilot = 80kg
The speed of plane v = 180km/h = 50m/s
The radius of circle r = 300 m
The acceleration of the plane will be calculated as:
a = v²/r
a = 50²/300
a = 2500/300
a = 8.33m/s²
Note that 180km/h was converted to m/s by
= (180 × 1000)/(60 × 60)
= 180000/3600
= 50m/s
1000 meters = 1 kilometer
60 minutes = 1 hour
60 seconds = 1 minute
3600 seconds = 1 hour
Since substances break down much faster under aerobic conditions, as oxygen helps break the molecules apart, in a process called oxidation, biodegradation that actually takes place does so slowly that effects may not be visible in decades after a product was buried, if even then.
Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If instead, you throw it downward, its downward acceleration after release is 9.8 m/s2.
Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities and accelerations. The direction of the net force acting on an object determines the direction of its acceleration. Uniform acceleration, non-uniform acceleration, and average acceleration are the three different forms of accelerated motions.
A free-falling object experiences a downward acceleration of 9.8 m/s/s (on Earth). This specific designation is given to the numerical value for an object in free fall because it is such an essential value. The longer an object is in free fall, the faster it descends toward the ground due to gravity. In actuality, an object's velocity rises by 9.8 m/s2, so it reaches 9.8 m/s by the time it begins to fall.
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