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3 years ago

How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?

Physics

1 answer:

Sonja [21]3 years ago

6 0

Answer:

Explanation: We have seasons because Earth's axis – the imaginary line that goes through the Earth and around which the Earth spins — is tilted. It's tilted about 23.5 degrees relative to our plane of orbit (the ecliptic) around the Sun. As we orbit our Sun, our axis always points to the same fixed location in space. Our northern axis points almost directly toward Polaris, the North Star.

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Why has the model of the atom changed over time?

Makovka662 [10]

<span>A.
The model of the atom evolved as scientists have made new discoveries.</span>

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2 years ago

Which stage occurs just before ignition in an internal combustion engine?

ddd [48]

I think it is this because compression stroke it needs to be compressed then open up when started.

5 0

3 years ago

Read 2 more answers

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1

andriy [413]

Answer:

$\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.$

Explanation:

<u>Given:</u>

Mass, $\rm m\pm\Delta m = 1.3\pm 0.4\ kg.$

Velocity, $\rm v\pm \Delta v = 5.2\pm 0.2\ m/s.$

where,

$\rm \Delta m,\ \Delta v$ are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

$\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.$

The uncertainty in kinetic energy is given as:

$\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.$