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3 years ago

Which natural phenomenon occurs as a result of earth rotating on its axis?

1 answer:

5 0

Seasons because the tilt of the Earth's axis causes the sun to hit different spots with more or less light. So if the Northern part of the Earth is tilted away from the Sun, it's winter. But if it's tilted towards the sun, it's summer. Hope that helps!

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W=225 J P=25 W t=? what's T

Mama L [17]

Explanation:

p=w/t

t=w/p

t=225/25=9seconds

5 0

2 years ago

The current through a 10 ohm resistor is 1.2 amperes.What is the potential difference across the resistor?

Natalija [7]

Answer: The potential difference across the resistor is 12 volts.

Explanation:

To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.

Mathematically,

$V\propto I\\V=IR$

Where,

V = potential difference = ?V

I = Current flowing = 1.2 A

R = Resistor = $10\Omega$

Putting values in above equation, we get:

$V=1.2\times 10=12V$

Hence, the potential difference across the resistor is 12 volts

6 0

3 years ago

A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the

Alecsey [184]

Answer:

Part a)

$KE = 77.95 J$

Part b)

$L = 3.16 m$

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

$KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2$

$KE = \frac{7}{10} mv^2$

$KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)$

$KE = 77.95 J$

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

$mgLsin\theta = KE$

$\frac{42}{9.81}(9.81)L sin36 = 77.95$

$L = 3.16 m$

Part c)

by equation of energy conservation we know that

$\frac{7}{10}mv^2 = mgL sin\theta$

so here we can see that distance L is independent of the mass of the sphere

7 0

3 years ago

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
$a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm$

3 0

3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
$m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg$

7 0

3 years ago