A box is pushed toward the right across a classroom floor.The force of friction on the box is directed toward the

you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll

-BARSIC- [3]

Technically, it should roll forever.

6 0

3 years ago

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

3 years ago

PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball

bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

q₁ = 3C

q₂ = 14C

Distance between balls = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

F = $\frac{k q_{1} q_{2} }{r^{2} }$

where k = 9 x 10⁹Nm²/C²

q is the charges

r is the distance

Input the variables and solve;

F = $\frac{9 x 10^{9} x 3 x 14 }{9000}$ = 4.2 x 10⁷N

8 0

3 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

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

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

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

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago

The law of reflection states that if the angle of incidence is 19 degrees, the angle of reflection is ___ degrees.

Wewaii [24]

using the law of refraction, the incidence is equal to the reflection, but not refraction

4 0

3 years ago

Read 2 more answers