Answer:
force pushed the object = 10 × 5 = 50N
Answer:
Explanation:
The acceleration of a circular motion is given by
where 
is the angular velocity and 
is the radius.
Angular velocity is related to the period, T, by
Substitute into the previous formula.
This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.
Answer:
The first minimum would be observed at 41.57°
Explanation:
v = 340m/s = speed of sound
f = 610Hz
d = 0.840m
λ = ?
Mλ = wsinθ
m = mth order minima
λ = wavelength incident on the single slit
θ = angular position of the mth minima
But, λ = v / f
λ = 340 / 610 = 0.557m
θ = sin⁻(mλ/d)
θ = sin⁻ [(1 * 0.557) / 0.840]
θ = sin⁻ 0.6635
θ = 41.57°
The first minimum would be observed at 41.57°
Population size is the actual number of individuals in a population. Population density is a measurement of population size per unit area, i.e., population size divided by total land area. Abundance refers to the relative representation of a species in a particular ecosystem.
<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
