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3 years ago

What happens to light that strikes a piece of glass

1 answer:

6 0

If a ray of light hits the surface of a sheet of glass, some light will be reflected by the surface of the glass. However, much of the light will pass through the glass, because glass is transparent. ... This 'bending of a ray of light' when it passes from one substance into another substance is called refraction.

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What happens to the car if it is traveling in a circular path suddenly encounters ice?

pychu [463]

the car's friction is reduced to zero

the car's centripetal force drops to zero

the car continues in a straight path from the point at which it encountered the ice

8 0

3 years ago

Read 2 more answers

A car is traveling at a constant speed of 12m/s , when the driver accelerates the car reaches a speed of 26 m/s in 6s what is th

DaniilM [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (26 m/s) - (12 m/s) = 14 m/s

Time for the change = 6 s

Acceleration = (14 m/s) / (6 s)

Acceleration = (14/6) (m/s²)

<em>Acceleration = 2.33 m/s²</em>

7 0

4 years ago

(II) A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 m • N which accelerates

kozerog [31]

Answer:

$m = 23.3 kg$

Explanation:

As we know that it will have constant torque on it

so the acceleration of the ball will be constant so here we can say that we can use kinematics equation

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

$160(2\pi) = 0 + \frac{1}{2}\alpha (15^2)$

$320 \pi = 112.5 \alpha$

so we have

$\alpha = \frac{320\pi}{112.5}$

$\alpha = 8.94 rad/s^2$

now we know that

$\tau = I \alpha$

$10.8 = I(8.94)$

$I = 1.21 kg m^2$

so we know that

$I = \frac{2}{5}mR^2$

here we know that

diameter = 0.72 m

so radius (R) = 0.36 m

$\frac{2}{5}m(0.36^2) = 1.21$

$m = 23.3 kg$

8 0

3 years ago

Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied

liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

$EF_x = 0$ ; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

5 0

3 years ago

A mass on a spring will oscillate vertically when it is lifted and released how will the values of elastic potential energy and

Rama09 [41]

The answer to this is

<span>A mass on a spring will oscillate vertically when it is lifted and released. The gravitational potential energy increases from a minimum at the lowest point to a maximum at the highest point. ... The total mechanical energy at those points is the sum of the elastic and gravitational potential energies.</span>

Hope This Helps!
:D

6 0

4 years ago