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2 years ago

Which is not correct when describing the formation of rainbows?

1 answer:

7 0

D- All wave lengths refract at the same angle

Explanation- The amount of refraction increases as the wavelength of light decreases. Shorter wavelengths of light are slowed more and consequently experience more bending than the longer wavelengths.

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In which of the following situation is work being done?

Afina-wow [57]

A wagon is pulled at an angle of 30 degrees to the horizontal.

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3 years ago

Help help help HELP AAAAA

charle [14.2K]

The answer is B

As seen on the graph, the bus maintains a 9m/s speed for a majority of the trip to school.

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3 years ago

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =16321.0769 N/C

The electric field strength at point (x,y) = (0cm, 20 mm) is =35321.58999 N/C

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=16321.0769 N/C

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=35321.58999 N/C

8 0

2 years ago

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

$v=\dfrac{d}{t}$

Where. d = distance

t = time

Put the value into the formula

$v=\dfrac{18.288}{2}$

$v=10\ m/s$ during that run

(b). We need to calculate their velocity

Using formula of speed

$v=\dfrac{\Delta d}{\Delta t}$

Put the value into the formula

$v=\dfrac{22.86-36.58}{2}$

$v=-6.86\ m/s$

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

$d=vt$

Put the value into the formula

$d=6.86\times1.3$

$d=8.91\ m$

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

8 0

3 years ago

A football player running for a touchdown is what energy

melamori03 [73]

Potential and kinetic energy are the two types of energy, but they do get separated into subgroups, for which I do not know. Hope that helps.

4 0

3 years ago