Answer:
if it were to be digital then you would be able to carry it around where ever and look at it to remind you of the memories, however it probably wouldnt last as long as the physical one. eg if you phone breaks there is a chance to restore the photo but you would had to have backed it up beforehand.
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:
D
Explanation:
If two unequal forces act in opposite directions, one larger force must cancel out the smaller force,
leaving the net force to be some number in one direction.
Take for example a game of tug-of-war; there are two OPPOSITE forces (groups of people) acting on the rope, one force is pulling with a force in the negative direction while the other force is pulling in the positive direction.
If the forces on the rope were unequal, then the stronger force (group) will pull everything in their direction.
The same will happen on two unequal forces of opposite directions acting on a wooden block. Therefore, since the resultant force will have a non-zero magnitude and direction, there will be a change in the block's motion and position.
Answer:
0.56 m/s
Explanation:
The speed of the head at the end of the interval in each case is the area under the acceleration curve. Then the difference in speeds is the difference in areas.
We can find the area geometrically, using formulas for the area of a triangle and of a trapezoid.
A = 1/2bh . . . . area of a triangle
A = 1/2(b1 +b2)h . . . . area of a trapezoid
If h(t) is the acceleration at time t for a helmeted head, the area under that curve will be (in units of mm/s) ...
Vh = 1/2(h(3)·3) +1/2(h(3) +h(4))·1 +1/2(h(4) +h(6))·2 +1/2(h(6))·1
Vh = 1/2(4h(3) +3h(4) +3h(6)) = 1/2(4·40 +3·40 +3·80) = 260 . . . mm/s
If b(t) is the acceleration for a bare head, the area under that curve in the same units is ...
Vb = 1/2(b(2)·2 +1/2(b(2) +b(4))·2 +1/2(b(4) +b(6))·2 +1/2(b(6)·1)
Vb = 1/2(4b(2) +4b(4) +3b(6)) = 1/2(4·120 +4·140 +3·200) = 820 . . . mm/s
Then the difference in speed between the bare head and the helmeted head is ... (0.820 -0.260) m/s = 0.560 m/s
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