False
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Answer:
1411.8 N/m
Explanation:
From Hooke's law;
F= Ke
Where
F= force on the spring
K= force constant
e = extension
But e= 8.50 × 10^-2m
F= weight = 12.0 kg × 10 = 120 N
K = F/e = 120/8.50 × 10^-2
K= 1411.8 N/m
Answer:
U = 9.1 m/s
Explanation:
from the question we are given the following
time (t) = 1.8 s
angle = 23 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below
v = u + 0.5at
- The final velocity (v) is zero since the ball comes to rest
- The time (t) it takes to get to the maximum height would be half the time it is in the air, t = 0.5 x 1.8 = 0.9
therefore
0 = u - (0.5 x 9.8 x 0.9)
u = 7.9 m/s
for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore
U sin 60 = 7.9
U = 7.9 ÷ sin 60
U = 9.1 m/s
Answer:
No, it is not necessary for them to have same mass.
Explanation:
Let both bodies have a density d1 and d2 respectively.
Since their volumes are equal V1 = V2
we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D
Hence, d1 = and d2 =
Taking the ratio of densities,we get
This implies that unless the bodies have same densities, the mass of the two bodies will not be same.
Answer:
C. 28.09 amu
Explanation:
The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abundance) and X-30 (29.974 amu, 3.10% abundance).
The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.
The atomic weight is computed as follows:
atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of X-30 × fractional abundance
atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 × 0.0310
atomic weight = 25.8031871 + 1.3531792 + 0.929194
atomic weight = 28.0855603 amu
To 2 decimal place atomic weight = 28.09 amu
