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3 years ago

6

Comparing infrared and ultraviolet, we can say that . Comparing infrared and ultraviolet, we can say that .

ultraviolet has longer wavelength and lower photon energy ultraviolet has longer wavelength and higher photon energy infrared has longer wavelength and lower photon energy infrared has longer wavelength and higher photon energy

1 answer:

Fudgin [204]3 years ago

8 0

Answer: Comparing infrared and ultraviolet, ''we can say that infrared has longer wavelength and lower photon energy''

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a 10kg cement block horizontally at 6 m/s plows into a bank of sand and comes to a stop in 2.0 m/s. What is the average impact f

Gnesinka [82]

I assume the block plows into the bank of sand with a velocity of 6 m/s and comes to a stop in 2 s.

$\bar Ft = \Delta mv \\ \\ \bar F = \frac{\Delta mv }{t} \\ \\ \bar F \ avarage \ Force \\ m \ momentum \\ v \ velocity \\ t \ time$

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3 years ago

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to ro

Dafna11 [192]

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

$T_x = nsinA = \frac{mv^2}{r}$

$T_y = ncosA = mg$

Dividing both.

$tan A = \frac{v^2}{rg}$

$tan A = \frac{11.7^2}{50*9.8}$

$A = tan^{-1} (0.279367)$

$A = 15.608\°$

Therefore the angle that should the curve be banked is 15.608°

7 0

4 years ago

At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A

saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B= $0.6\times 10^{-4} T$

$\theta=75^{\circ}$

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F= $IBlsin\theta$

Using the formula

$F=0.6\times 10^{-4}\times 15\times 19sin75$

$F=16.5\times 10^{-3} N$

Direction= $tan\theta=cot(90-75)=tan15^{\circ}$

$\theta=15^{\circ}$

15 degree above the horizontal in the northward direction.

5 0

3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 100 m in 25 s?

larisa86 [58]

As per the question, the distance travelled by bobsled [s] = 100 m

The time taken by the bobsled to travel that distance [t] = 25 s

We are asked to calculate the speed of the bobsled.

The speed of the bobsled is calculated as -

$speed\ =\ \frac{distance}{time}$

$v\ =\ \frac{s}{t}$

$=\frac{100\ m}{25\ s}$

$=\ 4\ m/s$

Hence, the correct answer to the question is A. 4 m/s.

8 0

3 years ago