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sergij07 [2.7K]

3 years ago

13

How is color related to wavelength?

1 answer:

dlinn [17]3 years ago

8 0

<span>Each color has a different wavelength allowing the eye to see it.</span>

You might be interested in

Balance the following chemical equation:<br> H3PO4 + HCl → PC15 + H20

kipiarov [429]

Answer:

H3PO4 + 5HCl → PCl5 + 4H2O

Explanation:

3 0

3 years ago

A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

trasher [3.6K]

Answer:

$\boxed {\boxed {\sf 500 \ Newtons }}$

Explanation:

The equation for force is given:

$F=m*a$

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

$m= 100 \ kg \\a= 5 \ m/s^2$

Substitute the values into the formula.

$F=100 \ kg * 5 \ m/s^2$

Multiply.

$F= 500 \ kg*m/s^2$

1 kilograms meter per square second is equal to 1 Newton.
Our answer of 500 kg*m/s² is equal to 500 Newtons.

$F= 500 \ N$

The force exerted by the go cart engine is <u>500 Newtons.</u>

4 0

3 years ago

Read 2 more answers

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

Sidana [21]

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

4 0

4 years ago

An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of

Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$ (taking the value of $v_g$ and $v_g$ at 200°C )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

$=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

$=850.65+3.377\times10^{-3}\times 1744.65$

= 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

= 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

= 1.51 kW

6 0

3 years ago

If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by

lina2011 [118]

Answer:

The value is $v = 1.309*10^{5}\ m/s$

Explanation:

The radius is $r = 7.40 *10^{20} \ m$

The mass of the ordinary matter is $M_{rod} = 1.90 *10^{41}\ kg$

Generally the speed of the star is mathematically represented as

$v = \sqrt{\frac{G * M}{r} }$

Here G is the gravitational constant with a value

$G = 6.67384 * 10^{-11}$

So

$v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }$

=> $v = 1.309*10^{5}\ m/s$

8 0

3 years ago