Answer:42.4m/s^2
Explanation:
Velocity(v)=6m/s
Radius(r)=0.85 meter
Centripetal acceleration=(v x v) ➗ r
Centripetal acceleration=(6 x 6) ➗ 0.85
Centripetal acceleration=36 ➗ 0.85
Centripetal acceleration=42.4
Answer:
The height is : 60.025 m
Explanation:
The flowerpot falls off the balcony with zero launch angle
Given the time of fright as 3.5 s then ;
The formula to apply is ;
3.5²= 2H/9.8
12.25 =2H/9.8
12.25 * 9.8 = 2H
120.05 = 2H
120.05/2 = H
60.025 =H
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Answer:
1170 m
Explanation:
Given:
a = 3.30 m/s²
v₀ = 0 m/s
v = 88.0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(88.0 m/s)² = (0 m/s)² + 2 (3.30 m/s²) (x - 0 m)
x = 1173.33 m
Rounded to 3 sig-figs, the runway must be at least 1170 meters long.
Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3
