Answer:
one-third of its weight on Earth's surface
Explanation:
Weight of an object is = W = m*g
Gravity on Earth = g₁ = 9.8 m/s
Gravity on Mars = g₂ = 
g₁
Weight of probe on earth = w₁ = m * g₁
Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )
As g₂ = g₁/3 --------- ( 2 )
Put equation (2) in equation (1)
so
Weight of probe on Mars = w₂ = m * g₁ /3
Weight of probe on Mars = m * g₁ = w₁
⇒Weight of probe on Mars = Weight of probe on earth
F(in) x d(in)= F(out)x d(out)
A machine has a mechanical advantage of 5.
F(out )= 5xF(in)300 newtons
F(in) = 300 N
F(out)= 5xF(in) = 5x300 N
= 1500 N
Work in = Work out = 3000 N x m3000
N x m = F_in x d_ind_in = 3000 Nxm / F_in = 3000 N x m / 300 N = 10 m3000 Nxm = F_out x d_out d_out = 3000 Nx m / F_out = 3000 N x m / 1500 N = 2 m
Answer:
-20.0 m/s and 30.0 m/s
Explanation:
Momentum is conserved:
m (30.0) + m (-20.0) = m v₁ + m v₂
30.0 − 20.0 = v₁ + v₂
10.0 = v₁ + v₂
Since the collision is perfectly elastic, energy is also conserved. Since there's no rotational energy or work done by friction, the initial kinetic energy equals the final kinetic energy.
½ m (30.0)² + ½ m (-20.0)² = ½ mv₁² + ½ mv₂²
(30.0)² + (-20.0)² = v₁² + v₂²
1300 = v₁² + v₂²
We now have two equations and two variables. Solve the system of equations using substitution:
1300 = v₁² + (10 − v₁)²
1300 = v₁² + 100 − 20v₁ + v₁²
0 = 2v₁² − 20v₁ − 1200
0 = v₁² − 10v₁ − 600
0 = (v₁ + 20) (v₁ − 30)
v₁ = -20, 30
If v₁ = -20, v₂ = 30.
If v₁ = 30, v₂ = -20.
So either way, the final velocities are -20.0 m/s and 30.0 m/s.
Answer : The atomic weight of the element is, 121.75 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
As we are given that,
Mass of isotope 1 = 120.9038 amu
Percentage abundance of isotope 1 = 57.25 %
Fractional abundance of isotope 1 = 0.5725
Mass of isotope 2 = 122.8831 amu
Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %
Fractional abundance of isotope 2 = 0.4275
Now put all the given values in above formula, we get:
![\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20element%7D%3D%5Csum%5B%28120.9038%5Ctimes%200.5725%29%2B%28122.8831%5Ctimes%200.4275%29%5D)
Therefore, the atomic weight of the element is, 121.75 amu
