Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
Answer:
you could measure several properties of
the unknown liquid and compare them with the properties of known
substances. You might observe and measure such properties as color,
odor, texture, density, boiling point, and freezing point.
Answer:
B
Explanation:
Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.
ΣF = 7 + 18 + (-20) = 5N to the right
At stp conditions (

), the speed of sound is

The sound wave moves by uniform motion, so we can use the basic relationship between space, time and velocity:

where S is the distance covered by the sound wave in a time t. In our problem, t=3.00 s, therefore the distance covered by the sound wave is
Answer:
F=50kg
D=5m
W=FD
therefore work done is 250Nm