A car initially traveling at 20 m/s accelerates at a uniform rate of 4.00 m/sec^2 for a displacement of 50 m
1 answer:
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1. A-20 km south east
The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:
east
south
Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:
and the direction is between the two original directions, so south-east.
2. D. 10 m/s
First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have
From which we find the total time of the fall, t:
Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is
3. B=32.32 m
As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have
And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is
So, the closest answer is B.
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