Ask question

Ask question

All categories

GenaCL600 [577]

3 years ago

13

Which of the following is NOT a skill scientists use to learn about the world?

1 answer:

sdas [7]3 years ago

8 0

There are many skills that scientists don't use to learn about the world.
These skills include playing the harmonica, riding a unicycle, and asking
others to choose from a list and then not telling them what's on the list.

You might be interested in

Why are there only two elements in the first period of the periodic table?(1 point)

Katen [24]

Answer:

because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits

6 0

2 years ago

A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the

Natali [406]

There are three forces acting on the book.
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.

4 0

3 years ago

A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0

3 years ago

Help please!!!!!!!!!!!!

Elena-2011 [213]

A would be most useful (Carbon dating)

3 0

3 years ago

What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?

Archy [21]

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

$I=\dfrac{E^2}{2\mu_{0}c}$

$E=\sqrt{2\times I\mu_{0}c}$

Put the value into the formula

$E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}$

$E=931.17\ N/C$

Using formula of magnetic field

$B = \dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{931.17}{3\times10^{8}}$

$B=0.0000031039\ T$

$B=3.10\times10^{-6}\ T$

(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using gravitational force

$F=\dfrac{Gm_{s}M_{e}}{r^2}$

Put the value into the formula

$F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}$

$F=3.522\times10^{22}\ N$

We need to calculate the radiation force

Using formula of force

$F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}$

Put the value into the formula

$F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2$

$F_{R}=4.8\times10^{8}\ N$

The gravitational and solar electromagnetic pressure forces of the sun on the earth

$\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}$

$\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}$

Hence, This is the required solution.

3 0

3 years ago