A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time . Thus, we can derive:

.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

.

Using this relationship in the distance formula for sound allows us to write:

.

Substituting the value of d from the first equation yields:

.

Now, after some calculations, we can proceed further:

.

Finally, the value is inserted into the initial equation:

Answer:

a)  τ = i ^ (y - z ) + j ^ (z Fₓ - x ) + k ^ (x - y Fₓ)

b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

Explanation:

a) The torque can be expressed as

        τ = r x F

To tackle this equation, using the determinant approach is the most straightforward method

         

The resulting expression is

      τ = i ^ (y - z ) + j ^ (z Fₓ - x ) + k ^ (x - y Fₓ)

b) Now let's compute

     τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)

     τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)

     τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) To find angular acceleration, we use

       τ = I α

       α = τ / I

The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.

           

     α = (-0.189i^  -5.6 j^  + 33.978k^) / 241

     α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

a) 3.56 x 10^22 N. b) 3.56 x 10^22 N. The sun’s mass is M = 2 x 10^30 kg, while the Earth's mass is m = 6 x 10^24 kg, with a distance of R = 1.5 x 10^11 m separating them. Applying Newton's law for gravitational force F = G (mM / R²), where G = 6.67 × 10^-11 m^3 kg^-1 s^-2 gives us F = 3.56 x 10^22 N. A) The gravitational force by the sun on Earth equates to the force exerted by Earth on the sun, which is also 3.56 x 10^22 N.

Answer:

The number of photons emitted each second is

Explanation:

Let 'n' stand for the quantity of photons released by the bulb.

Provided Information:

The bulb radiates energy at a rate of 100 J per second (E).

Wavelength of emitted light is (λ) = 525 nm =

The energy of a photon is calculated by:

Where,

Now, if we have 'n' photons, the total energy is equivalent to the energy of a single photon multiplied by the count of photons. Thus,

To express in terms of 'n', we find:

Insert the provided values and solve for 'n'. The resulting calculation yields

Consequently,

photons are discharged every second.

Ans    specifically, 4

Explanation: