Answer:
Jari
Explanation:
The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.
From Jari's line, the starting point is (0, 0) and another point is (6, 7)
The gradient being change in y to change in x
Change in y=7-0=7
Change in x=6-0=6
Slope is 7/6
For Jade, first point is (0, 10) then another point is (6, 16)
Change in y=16-10=6
Change in x=6-0=6
Slope is 6/6=1
Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade
Answer: E= KQ/r^2
Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.
The magnitude of an electric field E, at a point is given by Coulombs law as
E/ F/q
Where F= Coulombs force exertedon the charge and q= electric charge
E= F/q=(KQq)/r^2q
E=KQ/r^2
Answer:
t = 2 s
Explanation:
As we know that fish is pulled upwards with uniform maximum acceleration
then we will have

here we know that maximum possible acceleration of so that string will not break is given as

now we have


now for such acceleration we can use kinematics


t = 2 s
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11