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Sia

单鑫宇's Question
Physics M
Posted 4 months ago

Question 1/40
(2 p.)
A student throws a stone with velocity yy at an angle θ\theta to the vertical from the surface of a lake Air Iesistance can be ignored. The acceleration due to grovity is g\mathrm{g}.

What is the sine taken for the stone to ht the surface of the take?
vsinθg
\frac{v \sin \theta}{g}

A.
BB.
vcosθg
\frac{v \cos \theta}{g}

BB.
2vcosθ9
\frac{2 v \cos \theta}{9}

C.
D.
2vsinθg
\frac{2 v \sin \theta}{g}


Question 2/40 (2 p.)
A heavy ball falls freely, starting from rest. Between the third and fourth second of time it travels a distance of:
A. 4.9 m4.9 \mathrm{~m}
B. 9.8 m9.8 \mathrm{~m}
C. 29.4 m29.4 \mathrm{~m}
D. 34.3 m34.3 \mathrm{~m}
E. 39.8 m39.8 \mathrm{~m}
Question 3/43 / 4
Which of the fol moving with a c\mathrm{c}
xx
xx \mid \qquad
A
xx
I \qquad
D
A. A
B. B
C. C
D. D

Question 4/4
A stone thrown
A. circular
B. made of tt
C. byperboli
D. perabolio
E. a straight

Question 5/40
A 1000kg1000-\mathrm{kg} elevalu on the elevator is:
A. 6800 N6800 \mathrm{~N}
B. 1000 N1000 \mathrm{~N}
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Answer from Sia
Posted 4 months ago
Solution
a
The stone is thrown with an initial velocity vv at an angle θ\theta to the vertical. The vertical component of the initial velocity is vsinθv \sin \theta
b
The time taken for the stone to hit the surface of the lake can be found using the kinematic equation for vertical motion under gravity: t=vsinθg t = \frac{v \sin \theta}{g}
Answer
D
Key Concept
Kinematics of projectile motion: When an object is projected at an angle, its motion can be analyzed by resolving it into horizontal and vertical components. The vertical motion is influenced by gravity, and the time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity. Equation: t=vsinθg t = \frac{v \sin \theta}{g}
Explanation
The time taken for the stone to hit the surface of the lake is determined by the vertical component of its initial velocity and the acceleration due to gravity. Using the equation for vertical motion, we find that the time is given by t=vsinθg t = \frac{v \sin \theta}{g} , which corresponds to option D.

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