Answer:
= 7.07 m
Explanation:
The Tarzan reaches bottom of swing after descending 2.5 m,
change in his potential energy equals his kinetic energy at bottom of swing
m g h = (1/2) m v² ,
hence speed v of Tarzan at bottom of swing is given as
v = ( 2 g h )1/2
= ( 2 × 9.8 × 2.5 )1/2
= 7 m/s
At the bottom of swing, if the vine breaks, then he is moving with horizontal velocity 7 m/s in gravitational field.
If vertical distance from ground to bottom of swing is 5 m, then time t for Tarzan to reach ground is given by
S = (1/2)g t2 or t = (2S/g)1/2
= ( 2 × 5 / 9.8 )1/2
= 1.01 s
Horizontal distance traveled by Tarzan = 1.01 × 7
= 7.07 m
Neither the speed nor the distance of a falling object is linearly related to time.
Answer:
x = 42 m
y = -28 m
vx = 17.5 m/s
vy = -23.5 m/s
Explanation:
if the throw point is the origin and UP and AWAY are the positive directions.
s = s₀ + v₀t + ½at²
v = u + at
horizontal position = 0 + 17.5 m/s(2.4 s) + ½(0)(2.4)² = 42 m
vertical position = 0 + 0(2.4) + ½(-9.8 m/s²)(2.4 s)² = -28 m
horizontal velocity = 17.5 + 0(2.4) = 17.5 m/s
vertical velocity = 0 + (-9.8)(2.4) = -23.5 m/s
Answer:
a) F_{e} - F_{m} = 0, b) v = 666.67 m / s
Explanation:
For the proton to move y-axis the sum of the electric and magnetic force must be zero, therefore
= 0
a) ∑ F = 0
F_{e} - F_{m} = 0
b) we write the forces
q E = q v B
v = E / B
Let's calculate
v = 300 / 0.45
v = 666.67 m / s
Answer:
The atomic weight of the metal is 58.7 g/mol
Explanation:
Given;
density of the metal sample, ρ = 8.91 g/cm³
edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm
Volume of the unit cell of the metal;
V = α³
V = (352.4 x 10⁻¹⁰ cm)³
V = 4.376 x 10⁻²³ cm³
Mass of the metal in unit cell
mass = density x volume
mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³
mass = 3.899 x 10⁻²² g
Atomic weight, based on 4 atoms per unit cell;
4 atoms = 3.899 x 10⁻²² g
6.022 x 10²³ atoms = ?
= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)
= 58.699 g/mol
= 58.7 g/mol (this metal is Nickel)
Theerefore, the atomic weight of the metal is 58.7 g/mol