Answer:
682.32 m
Explanation:
Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s
time after which police man starts = 2 seconds
Acceleration of the policeman = 4.0 m/s²
let the time taken to catch be 't' seconds
Now,
The total distance to be covered by the policeman
= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t
thus,
From Newton's equation of motion
where,
s is the total distance traveled by the police man
u is the initial speed = 0
a is the acceleration
t is the time
thus,
= 33.33 × 2 + 33.33 × t
or
2t² = 66.66 + 33.33t
or
2t² - 33.33t - 66.66 = 0
on solving the above equation, we get
t = 18.47 seconds (negative value is ignored as time cannot be negative)
therefore,
the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m
C) When both objects have the same temperature.
<em>Hope this helps!</em>
Answer:
Explanation:
Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.
This can be easily seen if we use Gauss's law,
We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,
which is the same as that of a point charge.
In our problem the charges being of opposite signs, the electric field will add up. Therefore,
where, = distance between the center of one sphere to the midpoint (between the 2 spheres)
Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
Velocity is a speed AND its direction.
30 mph south
30 mph east
These are the same speed but different velocities.