Answer:
<h3>38,673.9N</h3>
Explanation:
According to newton's second law:
Force = mass * acceleration
Given
Mass = 873kg
acceleration = 44.66m/s²
Magnitude of the force is expressed as;
F = ma
F = 873 * 44.6
F = 38,673.9N
<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>
The largest transition metal is copernicium with 112 protons.
Answer:
19.53 cm
Explanation:
The computation of the height is as follows:
Here we applied the conservation of the energy formula
As we know that
P.E of the block = P.E of the spring
m g h = ( 1 ÷ 2) k x^2
where
m = 0.15
g = 9.81
k = 420
x = 0.037
So now put the values to the above formula
(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2
1.4715 (h) = 0.28749
h = 0.19537 m
= 19.53 cm
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
