You need to draw two lines from the head of the arrow: one parallel to the axis (which will be bent) and the other one passing through the center of the converging lens <span>(which will continue straight) </span>and see where they intersect.
If the arrow is at an infinite distance, the image will be point-like in F.
If the arrow is at a distance greater than 2F, the image will be upside down between F and 2F.
If the arrow is at a distance of 2F, the image will be upside down in 2F.
If the arrow is at a distance between F and 2F, the image will be upside down beyond 2F.
If the arrow is at a distance F, all rays proceed parallel to each other and no image is formed.
If the arrow is at a distance smaller than F, the image will be virtual.
Therefore, the correct matches are 1B, 2C, 3D, and 4A.
0.289 day because 25000 is in seconds 0.28935185 Days hope that help!
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes6.88%5Ctimes10%5E%7B-6%7D%7D%7B%283.00%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=6.88\times10^{7}\ N/C](https://tex.z-dn.net/?f=E%3D6.88%5Ctimes10%5E%7B7%7D%5C%20N%2FC)
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
Put the value into the formula
![E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes%28-2.96%5Ctimes10%5E%7B-6%7D%29%7D%7B%287.00%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![E=5.43\times10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D5.43%5Ctimes10%5E%7B6%7D%5C%20N%2FC)
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.
Explanation:
Momentum is mass times speed.
p = mv
a) p = (1500 kg) (25.0 m/s) = 37,500 kg m/s
b) p = (40,000 kg) (1.00 m/s) = 40,000 kg m/s
The truck has more linear momentum.
Momentum in the y direction:
pᵧ = (1500 kg) (25.0 m/s) = 37,500 kg m/s
Momentum in the x direction:
pₓ = (1500 kg) (15.0 m/s) = 22,500 kg m/s
Total linear momentum:
p² = pₓ² + pᵧ²
p² = (22,500 kg m/s)² + (37,500 kg m/s)²
p = 43,700 kg m/s
Answer:
Mariner 10 in 1974 and 1975
Explanation: