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3 years ago

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A museum sets up a display of fluorescent minerals. Which best describes how electromagnetic waves can be used to enhance the di

splay?
A. Infrared light shining on the minerals will cause them to give off visible light.
B. Visible light shining on the minerals will cause them to give off infrared light.
C. Ultraviolet light shining on the minerals will cause them to give off visible light.
D. Visible light shining on the minerals will cause them to give off ultraviolet light.

2 answers:

belka [17]3 years ago

8 0

Answer:

C.

Explanation:

Ultraviolet light shining on the minerals will cause them to give off visible light.

Nata [24]3 years ago

3 0

The answer is C.
Pick C because i am right .. ty

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The __________-second rule applies to any speed in ideal weather and road conditions.

SOVA2 [1]

The two-second rule.

It is a common guideline to follow while driving.

It means that any given driver should be AT LEAST two seconds behind any vehicle that is driving in front of his vehicle. It might apply for any kind of vehicle.

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3 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

Read 2 more answers

Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m

Rufina [12.5K]

<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

<em>But;</em>

A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

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3 years ago

The football player running toward the goal line has

soldi70 [24.7K]

Kinetic energy.

Kinetic energy is the type of energy observed in moving objects. In this case the football player is running, ie moving, so he/she must have kinetic energy.

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2 years ago