Is the system in the image moving? Explain. If it is moving, find its acceleration.

Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h

andreyandreev [35.5K]

Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.

Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.

4 0

2 years ago

Read 2 more answers

Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j

choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

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8 0

2 years ago

Please give an answer that is coherent

Semmy [17]

Answer:

480

Explanation:

resistance equals to potential difference divide by electric current

120÷0.25

=480

6 0

3 years ago

. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa

lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

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(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

3 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago