If 320 J of work is done on a spring with a spring constant of 730 N/m, how far will it stretch?
1 answer:
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Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration
where
Vo = Initial speed =0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time
T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed
c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
Using the pressure law (P1 x V1)/ T1 = (P2 x V2)/ T2 where P1= the initial pressure V1= initial volume T1= initial temperature and P2= the final pressure V2= the final volume T2 = the final temperature and temperature is always in kelvin
Answer:
On moon time period will become 2.45 times of the time period on earth
Explanation:
Time period of simple pendulum is equal to ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth
As when we go to moon, acceleration due to gravity on moon is times os acceleration due to gravity on earth
So time period of pendulum on moon is equal to
--------eqn 2
Dividing eqn 2 by eqn 1
So on moon time period will become 2.45 times of the time period on earth