Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
- Weight (W) = 110 N
- Acceleration due to gravity (g) = 9.8 m/s^2
- Let the mass of the object be m.
- By using the formula, W = mg, we get,
- 110 N = 9.8 m/s^2 × m
- or, m = 110 N ÷ 9.8 m/s^2
- or, m = 11.2 Kg
<u>Answer:</u>
<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>
Hope you could get an idea from here.
Doubt clarification - use comment section.
The measuring devices for mass is scales and Balances
5Newtons or 5N
Ten newtons minus five Newton’s is 5 Newton’s