Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,
substituting the values in the equation we get,
f = 1.03 x 10⁸Hz
Now,
The time period (T) = 
or
T = 
= 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target = 
= 145 m
Answer:(a)360N,(b)171N,(c)2.702m
Explanation:
(a)Maximum Friction Force =
=360 N
(b)Moment about Ground Point
(c)
Here maximum friction force can be 360 N
Therefore 
Where x is the maximum distance moved by man along the ladder
740x=2000
x=2.702m
Answer:
B - A
Explanation:
For the combination of 2 vector to due southwest, 1 vector must due south and the other vector due west. Since vector B is already due west, vector A should due south. As vector A is already due north, vector -A would due south. So the combination of B + (-A) or B - A should points southwest
Answer:
The answer to your question is: a = 1.99 m/s²
Explanation:
Data
mass = 20 kg
angle = 56°
Force = 71 N
horizontal acceleration = ?
Process
Find the horizontal force
cos Ф = adjacent side / hypotenuse
adjacent side = hypotenuse x cosФ
adjacent side = 71 x cos 56
a.s. = 39.70 N
Newton's second law
F = ma
a = F/m
a = 39.7 / 20
a = 1.99 m/s²
Answer:
50N
Explanation:
Force (N) = mass (kg) × acceleration (m/s²)
0.25kg times 200m/s² = 50N
