Speed = wavelength × frequency
giving that frequency is 0, wavelength and speed are directionally proportional. wavelength decrease = speed decrease
Assuming the object is on earth the objects weight would be equal to its mass multiplied by the gravitational field constant
mass=22kg
g=9.80665N/kg
weight=(22 kg) (9.80665 N/kg)=215.7463N
generally g is rounded to be 10 N/kg so for any question where it asks the weight given the mass just multiply by 10 and that should suffice. In this case the answer would be 220 N
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
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
Answer
given,
heat added to the gas,Q = 3300 kcal
initial volume, V₁ = 13.7 m³
final volume, V₂ = 19.7 m³
atmospheric pressure, P = 1.013 x 10⁵ Pa
a) Work done by the gas
W = P Δ V
W = 1.013 x 10⁵ x (19.7 - 13.7)
W = 6.029 x 10⁵ J
b) internal energy of the gas = ?
now,
change in internal energy
Δ U = Q - W
Q = 3300 x 10³ cal
Q = 3300 x 10³ x 4.186 J
Q = 1.38 x 10⁷ J
now,
Δ U = 1.38 x 10⁷ - 6.029 x 10⁵
Δ U = 1.32 x 10⁷ J
