Answer:
5.3 m/s
Explanation:
First, find the time it takes for him to fall 7m.
y = y₀ + v₀ t + ½ at²
0 = 7 + (0) t + ½ (-9.8) t²
0 = 7 − 4.9 t²
t ≈ 1.20 s
Now find the velocity he needs to travel 6.3m in that time.
x = x₀ + v₀ t + ½ at²
6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²
v₀ ≈ 5.27 m/s
Rounded to two significant figures, the man must run with a speed of 5.3 m/s.
It is customary to work in SI units.
Calculate the volume of the concrete.
V = 3.7*2.1*5.8 cm³ = 45.066 cm³ = 45.066 x 10 ⁻⁶ m³
The mass is 43.8 g = 43.8 x 10⁻³ kg
The density is mass/volume.
Density = (43.8 x 10⁻³ kg)/(45.066 x 10⁻⁶ m³) = 971.9 kg/m³
Answer: 971.9 kg/m³
<u>We are Given:</u>
Mass of the block (m) = 500 grams or 0.5 Kg
Initial velocity of the block (u) = 0 m/s
Distance travelled by the block (s) = 8 m
Time taken to cover 8 m (t)= 2 seconds
Acceleration of the block (a) = a m/s²
<u>Solving for the acceleration:</u>
From the seconds equation of motion:
s = ut + 1/2* (at²)
<em>replacing the variables</em>
8 = (0)(2) + 1/2(a)(2)²
8 = 2a
a = 4 m/s²
Therefore, the acceleration of the block is 4 m/s²
Get to school and learn boi
Answer:
a) k = 2231.40 N/m
b) v = 0.491 m/s
Explanation:
Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.
when the box encounters the spring, all the energy of the box is kinetic energy:
the energy relationship between the box and the spring is given by:
1/2(m)×(v^2) = 1/2(k)×(x^2)
(m)×(v^2) = (k)×(x^2)
a) (m)×(v^2) = (k)×(x^2)
k = [(m)×(v^2)]/(x^2)
k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)
k = 2231.40 N/m
Therefore, the force spring constant is 2231.40 N/m
b) (m)×(v^2) = (k)×(x^2)
v^2 = [(k)(x^2)]/m
v = \sqrt{ [(k)(x^2)]/m}
v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}
= 0.491 m/s
