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2 years ago

Can lamp that works on a 2.5 v work on a 1.12 v ?

Physics

1 answer:

12345 [234]2 years ago

5 0

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

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A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

6 0

3 years ago

How are the magnetic domains of a magnet different from the domains of an ordinary piece of metal?

My name is Ann [436]

Answer:

In a magnet, the domains all point in the same direction; in an ordinary piece of metal, they're all jumbled up.

Explanation:

In a magnet, the domains all point toward the north pole; in an ordinary piece of metal, they all point to the south pole.

Side note:
Hope this helps!
Please give Brainliest!

6 0

2 years ago

Ok ok ik ik this is not the answer but I think you need to hear this

leonid [27]

Answer

oh thanksssss i hope u have a great day

Explanation:

ur a really awesome person and i thank u for that

4 0

2 years ago

a man uses a 900 watt electric iron to press his clothes for an average of 4 hours a week for five weeks if the cost of an elect

ser-zykov [4K]

Answer:

54 Kobo

Explanation:

Units of <u>electricity</u> are measured in kilowatt hours (kWh).

Given information:

900 watt electric iron
Appliance usage = 4 hours a week for 5 weeks
Unit cost of electricity = 3 Kobo per kWh

Convert the wattage of the electric iron from watts to kilowatts.

1000 watts (W) = 1 kilowatt (kW)

⇒ 900 watts = 1 ÷ 1000 = 0.9 kilowatts

This means that the power consumption of the electric iron is 0.9 kW per hour of use.

Total hours spent pressing clothes:

= 4 hours per week for 5 weeks

= 4 × 5

= 20 hours

<u>Total power consumption</u>:

= number of kW × number of hours

= 0.9 × 20

= 18 kWh

To find the <u>total cost</u>, multiply the total kWh by the cost per kWh:

⇒ Cost = 18 × 3 = 54 Kobo

4 0

2 years ago

Read 2 more answers

A mover loads a crate onto a truck bed 1.6m from the street using a ramp that is 4.6m long. What is a mechanical advantage?

Ne4ueva [31]

Answer:

Mechanical advantage = 2.875

Explanation:

Given:

A diagram is shown below for the above scenario.

Length of ramp (Effort arm) = 4.6 m

Height of truck bed ( Resistance length) = 1.6 m

Mechanical advantage (MA) is the ratio of effort arm and resistance length.

So, mechanical advantage is given as,

$MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875$

6 0

3 years ago

Can lamp that works on a 2.5 v work on a 1.12 v ?​

Can lamp that works on a 2.5 v work on a 1.12 v ?