Answer:
SECOND LAW OF NEWTON
Explanation:
When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression
I = F t = Δp
As we can see this expression is a form of Newton's second law
F = m a
a = dv / dt
F = m dv / dt
F dt = m dv
p = mv
F dt = dp
Therefore the station moves through the SECOND LAW OF NEWTON
Answer:
4.02 km/hr
Explanation:
5 km/hr = 1.39 m/s
The swimmer's speed relative to the ground must have the same direction as line AC.
The vertical component of the velocity is:
uᵧ = us cos 45
uᵧ = √2/2 us
The horizontal component of the velocity is:
uₓ = 1.39 − us sin 45
uₓ = 1.39 − √2/2 us
Writing a proportion:
uₓ / uᵧ = 121 / 159
(1.39 − √2/2 us) / (√2/2 us) = 121 / 159
Cross multiply and solve:
159 (1.39 − √2/2 us) = 121 (√2/2 us)
220.8 − 79.5√2 us = 60.5√2 us
220.8 = 140√2 us
us = 1.115
The swimmer's speed is 1.115 m/s, or 4.02 km/hr.
Answer:
Total displacement will be 47 meter
Total distance will be 83 meters
Explanation:
We have given that first the student go eastward towards bus stop 20 meters
But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters
So displacement = 20-18 = 2 meters
And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters
Total distance traveled by the student = 20+18+45 = 83 meters
Answer: 
Explanation:
The confidence interval for population mean is given by :-
Given : Sample size : 
Sample mean : 
Standard deviation : 
Significance level : 
Critical value : 
Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
