Answer:
The string must support the tension of 392 N.
Explanation:
The tension that the string must support should equal the centripetal force exerted on the on the stone as it goes in a circular path (because if the string supported less tension, it would break).
The centripetal force exerted on the stone is
where
<em>v</em> = velocity of the stone in m/s
<em>m</em> = mass of the stone in kg
<em>R</em> = radius of the circular path.
Now the velocity of the stone is 7.00 m/s, the mass of the stone is 4000g or 4 kg (1000 g = 1kg), and the radius of the circular path is just the length of the string, and it is 50 cm or 0.5 m (100cm =1m); therefore, we get
m = 4kg
v =7m/s
R = 0.5m.
We put these values into the equation for the centripetal force and get:
The centripetal force is 392 Newtons, and therefore, the tension that the string must support mus be 392 N.
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Answer is 1 molecule of S
Answer:
You need to put the diagrams. repost and include the diagrams
Answer:
9.8×104Nm−2,0.025,3.92×106Nm−2
Solution :
Here, L=10cm=10×10−2m
F=100kgf=100×9.8N
ΔL=0.25cm=0.25×10−2m,
Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025
=3.92×106Nm−2
Explanation: