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Alisiya [41]
4 years ago
5

The element in an incandescent light bulb that releases light energy is _____. a phosphor mercury vapor a thin tungsten filament

a thick tungsten filament
Physics
2 answers:
Ivenika [448]4 years ago
7 0

<u>Answer:</u> The correct answer is thin tungsten filament.

<u>Explanation:</u>

Incandescent light bulb is a substance which converts electrical energy to light and heat energy and has fine filament made of metal.

These bulbs have fine filament to have the maximum resistance as we know that resistance is inversely proportional to the area of cross section of the wire.

Metal used in the filament is tungsten because it has the highest melting point of 3422^oC

Hence, the correct answer is thin tungsten filament.

Dimas [21]4 years ago
6 0
The answer to this physics question would be 'a thin tungsten filament'. The element in an incandescent light bulb that releases light energy is a thin tungsten filament. All the other choices are not applicable as answers to this question.
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A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0
3 years ago
"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1
kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

\texttt{ }

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

\boxed{ \omega = \omega_o + \alpha t }

\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }

\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }

\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

\texttt{ }

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

\theta = \frac{( \omega + \omega_o )}{2} t

\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0

\theta = 10 \times 4.0

\boxed {\theta = 40 \texttt{ rad}}

\texttt{ }

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

\texttt{average angular speed} = \theta \div t

\texttt{average angular speed} = 40 \div 4.0

\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Moment of Inertia : brainly.com/question/13796477
  • The Ratio of the Moments of Inertia : brainly.com/question/2176655

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0
4 years ago
Read 2 more answers
A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i
Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2   Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft     and      d(5) = 35 ft

V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft     and      d(2.1) = 8.61 ft

V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s

4 0
4 years ago
Scientists on Earth are able to communicate with rovers on Mars through their receivers, which can receive and process signals.
Bas_tet [7]
B electromagnetic waves I think
4 0
3 years ago
Read 2 more answers
Curious George is whirling a 2.0 kg bunch of bananas on a smooth floor in a circular path having a radius of 0.50 m. What force
Sedaia [141]

Answer:

The force will be F_{c}=2.47 N

Explanation:

Let's use the centripetal force equation.

F_{c}=m\omega ^{2}R

Where:

m is the mass of the bunch of bananas

ω is the angular speed

R is the radius

Now, 1 rev every 4 seconds or 0.25 rev/sec is the angular speed, but we need to write this speed in rad per second.

\omega =0.25\frac{rev}{s}=0.25*2\pi \frac{rad}{s}=1.57 \frac{rad}{s}

FInally, the force will be:

F_{c}=2.0*1.57^{2}*0.5

F_{c}=2.0*1.57^{2}*0.5

F_{c}=2.47 N

I hope it helps you!

4 0
3 years ago
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