A 1.5m wire carries a 2 A current when a potential difference of 55 V is applied. What is the resistance of the wire?

Using an example, explain how being able to understand scientific principles and think scientifically can help you solve problem

crimeas [40]

One example is that when their food catches fire in a cooking pot or pan, it is advisable to smother the fire by covering the fire with a blanket rather than douse the fire with water. DOeing the later will only aggravate the fire. Knowledge of basic science dictates that oil and water are immiscible liquids the oil will continue to burn even as the water is poured into the oil. More so, as the water violently evaporates due to the heat, it spreads the burning oil and hence spreads the fire.

7 0

3 years ago

Read 2 more answers

The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colo

Bogdan [553]

Answer:

$4.58\times 10^{14}Hz$ $5.825\times 10^{14}Hz$ $6.31\times 10^{14}Hz$

Explanation:

We know that there is a relation between the speed of light ,frequency and wavelength.This relation can be given as $v=\nu \lambda$ where v is speed of light $\nu$ is frequency of light $\lambda$ is wavelength of light

In first case we have given λ=655 nm

so using formula $v=\nu \lambda$

$3\times 10^8=655\times 10^{-9}\times \nu$

$\nu =4.58\times 10^{14}$ Hz

In second case we have wavelength 515 nm so by using above formula

$3\times 10^8=515\times 10^{-9}\times \nu$

$\nu =5.825\times 10^{14}\ Hz$

In third case we have wavelength 475 nm

so $3\times 10^8=475\times 10^{-9}\times \nu$

$\nu =6.31\times 10^{14}Hz$

3 0

4 years ago

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0

3 years ago

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

4 years ago

A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu

neonofarm [45]

Answer:

$f = ma \\$