1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Liula [17]
3 years ago
14

The resistance of 100 W bulb is less than resistance of 40 W bulb. Explain the reason.

Physics
1 answer:
vazorg [7]3 years ago
4 0

Ok so we know that electric power is:

P=UI=I^2R

If we express resistance.

R=\dfrac{P}{I^2}

Now if you have 100W of power you will probably get a bigger resistance than with 40W of power.

However here it says that the resistance of 40W bulb is bigger than 100W bulb. Which means the statement is incorrect.

Hope this helps.

r3t40

You might be interested in
Why is it correct to say that the linear electron flow of the light reactions directly captures energy in nadph but only indirec
denis-greek [22]

Answer:

he electron is directly transferred to NADP+ to NADPH, but electron flow is used to generate a proton gradient for ATP synthesis. Electron is not directly transferred to ATP

NADP= Nicotin amide adenine dinucleotide phosphate

NADPH =Nicotinamide adenine dinucleotide phosphate is the reduced form of  NADP

therefore electron is not directly transferred to ATP

5 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after
worty [1.4K]
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t, 
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.
3 0
3 years ago
My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s
marishachu [46]

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

7 0
2 years ago
Which of the following gases are the heaviest? <br> O2, CH4, CO2, Cl2
kvasek [131]

the answer is

CI2 because its 70.


5 0
3 years ago
Other questions:
  • Which hemisphere is experiencing summer in july?
    5·1 answer
  • Make a general statement concerning how large bodies of water affect the climate of nearby coastal communities.
    5·1 answer
  • A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
    15·2 answers
  • a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun
    13·2 answers
  • A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.24 m/s2 for 10.0 s. 2
    13·1 answer
  • What type of speed looks at a particular point in time?
    8·1 answer
  • What does wave speed have to do with all SONAR AND RADAR technologies?
    12·2 answers
  • 93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
    11·1 answer
  • #1. What, roughly, is the percent uncertainty in
    11·1 answer
  • 3. A ball of mass 2.0 kg is attached to a vertical spring. The spring constant is 196 N/m. What is the period if it were to be i
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!