E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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Answer:
Hope i could help!
Explanation:
so all but one light could be burned out, and the last one will still function.
Answer:
0.266 m
Explanation:
Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,
P= mv where p is momentum, m is mass and v is the speed of an object. In this case
where sunscripts p and b represent putty and block respectively, c is common velocity.
Substituting the given values then
3*8=v(15+3)
V=24/18=1.33 m/s
The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy
where k is spring constant and x is the compression of spring. Substituting the given values then
