Remember Newton's second law: F=ma
to get the force in newtons, mass should be in kg and acceleration in m/s^2
conveniently, we don't need to convert units
we just need to multiply the two to get the force
65* 0.3 = 19.5 kg m/s^2 or N
if significant digit is an issue, the least number if sig figs is 1 so the answer would be 20 N
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x 
m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x 
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x x 3 x )/2
=120m
Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes
Now after applying brakes the distance covered by the car before it stops is given by kinematics equation
here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have
So total distance after which car will stop is
So car will not stop before the intersection as it is at distance 20 m
Answer;
-Tsunami
Explanation;
-Tsunami is a series of large ocean waves (or "wave train") of extremely long wavelength and period, usually generated when a gigantic body of water, such as an ocean, is suddenly displaced on a massive scale by an underwater disturbance such as an earthquake occurring on or near the sea floor or a volcanic eruption.
-After a sudden displacement of a large water volume by seismic activity (earthquake), the ocean floor is raised or dropped and large tsunami waves can be formed by gravitational forces.
Answer:
t = 39.60 s
Explanation:
Let's take a careful look at this interesting exercise.
In the first case the two motors apply the force in the same direction
F = m a₀
a₀ = F / m
with this acceleration it takes t = 28s to travel a distance, starting from rest
x = v₀ t + ½ a t²
x = ½ a₀ t²
t² = 2x / a₀
28² = 2x /a₀ (1)
in a second case the two motors apply perpendicular forces
we can analyze this situation as two independent movements, one in each direction
in the direction of axis a, there is a motor so its force is F/2
the acceleration on this axis is
a = F/2m
a = a₀ / 2
so if we use the distance equation
x = v₀ t + ½ a t²
as part of rest v₀ = 0
x = ½ (a₀ / 2) t²
let's clear the time
t² = (2x / a₀) 2
we substitute the let of equation 1
t² = 28² 2
t = 28 √2
t = 39.60 s
