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Serhud [2]
3 years ago
8

A green object will absorb ____________________ light and reflect ____________________ light. (ref: p.447-455)

Physics
1 answer:
Delvig [45]3 years ago
6 0

Answer:

A green object will absorb all light except for green light and reflect blue and yellow light.

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Recall that if a line is parallel to the vector v and passes through the point P0, which is the tip of the position vector r0, t
Elenna [48]

Answer:

The vector equation of the line is

                                \overrightarrow{r}=+t

Parametric equations for given line are

                                       x=7+t\\y=-8+6t\\z=3-13t

Explanation:

The vector equation of the line is given by

                                     r(t) = r_{o} + tv

r₀ = (7, -8, 3)

v = (1, 6, -13)

At these points the vector equation for this line is:

\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=+t

Parametric equations for given line are

                                       x=7+t\\y=-8+6t\\z=3-13t

8 0
3 years ago
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

4 0
3 years ago
the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force
-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

F_{net} = 51 N

now we also know that friction force due to ground is given as

F_f = 24 N

now we have

F_{net} = F_{ext} - F_f

51 = F_{ext} - 24

F_{ext} = 75 N

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

a = \frac{F_f}{m}

a = \frac{24}{24} = 1 m/s^2

now we can find the distance by kinematics

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2(-1)d

d = 1.125 m

so the distance moved will be 1.125 m

6 0
3 years ago
A boy throws a stone straight upward with an initial speed of 15m/s. What maximum height will the stone reach before falling bac
Vitek1552 [10]
height=\frac{velocity^2}{2gravity}\\\\
velocity=15\frac{m}{s}\\\\
g=9,8\frac{m}{s^2}\\\\height=\frac{15^2}{2*9,8}=\frac{225}{19,6}\\\\
\boxed{height=11,48\ meters}
6 0
3 years ago
An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =
goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

7 0
4 years ago
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