Help me with 1 and 2

What type of interaction allows the sidewalk to heat up and cook the egg?

Bogdan [553]

Answer:

Adsorption

Explanation:

Sidewalk cooking of egg is very popular in America. In summer people release fireworks in night sky and cook egg on the concrete sidewalk to check the level of temperature. When sunlight fall on the sidewalk most of the light is reflected back but some darker material adsorbs some photon, and when these photons are transferred to egg molecules it causes vibration among them and produce heat and cook egg.

7 0

3 years ago

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A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the ho

podryga [215]

<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s. To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So 0.50 s * 14.7447368 m/s = 7.372368399 m 1.00 s * 14.7447368 m/s = 14.7447368 m 1.50 s * 14.7447368 m/s = 22.1171052 m 2.00 s * 14.7447368 m/s = 29.48947359 m Rounding the results to 1 decimal place gives 0.50 s = 7.4 m 1.00 s = 14.7 m 1.50 s = 22.1 m 2.00 s = 29.5 m</span>

6 0

3 years ago

Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

brainly.com/question/1042017

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4 0

1 year ago

What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu

andreyandreev [35.5K]

Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

Explanation:

first wavelength, λ1 = 410 nm = 410 x 10^-9 m

Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

The relation between the energy and the wavelength is given by

E = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

h = 6.63 x 10^-34 Js

c = 3 x 10^8 m/s

So, energy correspond to first wavelength

E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J

E1 = 4.9 x 10^-19 J

So, energy correspond to second wavelength

E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J

E2 = 2.7 x 10^-19 J

4 0

3 years ago

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is: