Answer:
1.09 km, 3 km
Explanation:
displacement, d = 3.20 km at 20° North of east
A vector quantity has two components one along the x axis and the other is along Y axis. the component along X axis is called the horizontal component and it is due east.
The component along y axis is called vertical component and it is due North.
Displacement due North, dy = 3.20 Sin 20° = 1.09 km
Displacement due east, dx = 3.20 Cos 20° = 3 km
We use about 38 percent of Earth's land surface for agriculture
<span>k = 1.7 x 10^5 kg/s^2
Player mass = 69 kg
Hooke's law states
F = kX
where
F = Force
k = spring constant
X = deflection
So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration.
F = kX
F/X = k
115 kg* 9.8 m/s^2 / 0.65 cm
= 115 kg* 9.8 m/s^2 / 0.0065 m
= 1127 kg*m/s^2 / 0.0065 m
= 173384.6154 kg/s^2
Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2
Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So
X/0.39 cm = 115 kg/0.65 cm
X = 44.85 kg/0.65
X = 69 kg
The player masses 69 kg.</span>
Answer:
Let the weight of the person be W and be located at a distance 'a' from the left scale as shown in the figure
Since the body is in equilibrium we can use equations of statics to analyse the problem.
Taking Sum of Moments about A we have
Taking Sum of Moments about B we have
Solving the above 2 equations for W and 'a' we get
