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3 years ago

11

a moving electron accelerates at 5200m/s^2 in a 55.0 direction. After 0.350as,it has a velocity of 6598m/s in a -20.5 direction.

What is the y-component of the initial velocity?

1 answer:

Lady bird [3.3K]3 years ago

6 0

Answer:

-3802 m/s

Explanation:

The y-component of the final velocity is ...

(6598 m/s)·sin(-20.5°) ≈ -2310.7 m/s

The y-component of the velocity due to acceleration is ...

(5200 m/s²)(0.350 s)sin(55°) ≈ 1490.9 m/s

Then the initial velocity in the y-direction is found from ...

initial velocity + change in velocity = final velocity

initial velocity = (final velocity) - (change in velocity)

= -2310.7 m/s - 1490.9 m/s ≈ -3802 m/s

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3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago

When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be

Wittaler [7]

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

$V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]$

We now know that density is defined as the relationship between mass and volume.

$Ro = m/V_{d}$

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

$Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]$

7 0

2 years ago

A 3.00 μF capacitor is charged to 480 V and a 4.00 μF capacitor is charged to 500 V . Part A These capacitors are then disconnec

pogonyaev

Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1= 3×10^-6 F

V1= 480v

C2= 4×10^-6 F

V2= 500v

(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V

Simplifying the above, we get:

( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.

Further simplified as:

3440 × 10^-6 = 7 × 10^-6 × V

Making V the subject

V = 491.43volts

Therefore the potential difference across each capacitor is 491.43v

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3 years ago

The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f

Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

$X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}$

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon = 4 × 10⁸ m

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Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0

3 years ago