Answer:
Explanation:
Using apparent = mass of liquid expelled/mass of liquid remaining × temp. rise
Mass expelled: 250-248.5=1.5g
Mass remaining =248.5g
Temp. rise= 60-30= 30ºc
= 1.5/248.5×30=0.0002012
Using real= apparent + cubic expansivty
Cubic= 3 × linear expansivty
= 3×0.000006=0.000018
0.0002012+0.000018=0.0002192
Answer:
15.68 m/s
Explanation:
Given that,
She catches the ball 3.2 s later at the same height from which it was thrown.
When it reaches the maximum height, its height is equal to 0.
It will move under the action of gravity.
2 here comes for the time of ascent and descent.
So,
So, the initial upward speed of the ball is 15.68 m/s.
Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
The dimension of K is M/ T^2
according to the question T=2π square root ofm/k here 2 pi is constant so
T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2
complete question :
A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?
To learn more about dimension:
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Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
We solve for initial speed
Now, using the same expression we estimated time to first reach 18.5 m :
Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.