<h2>
Distance traveled in 1 second after drop is 4.9 m</h2><h2>
Distance traveled in 4 seconds after drop is 78.4 m</h2>
Explanation:
We have s = ut + 0.5at²
For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²
Substituting
s = 0 x t + 0.5 x 9.8 x t²
s = 4.9t²
We need to find distance traveled in 1 s and 4 s
Distance traveled in 1 second
s = 4.9 x 1² = 4.9 m
Distance traveled in 4 seconds
s = 4.9 x 4² = 78.4 m
Distance traveled in 1 second after drop = 4.9 m
Distance traveled in 4 seconds after drop = 78.4 m
Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
Answer:
14869817.395 m
Explanation:
=22 microarcsecond
λ = Wavelength = 1.3 mm
Converting to radians we get

From Rayleigh Criterion

Diameter of the effective primary objective is 14869817.395 m
It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.
Answer:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m
height of basket is 3.05 m
Launching angle

y=1.05
equation of trajectory of ball is given by

for x=12.27

u=10.69
for x=11.73

u=11.436 m/s
Thus range of speed is (10.69, 11.436)