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denis23 [38]
1 year ago
7

What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

electric field of magnitude 590 N/C
Physics
1 answer:
Pani-rosa [81]1 year ago
8 0

     charge must be equal to 5.74 ×10⁻⁵

 In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

     →    Fnet =0

     →    mg =  qE

 substituting the values we get :

         0.00345 × 9.81 =  q × 590

   →       q = 5.74 ×10⁻⁵

    Hence the charge must be equal to   5.74 ×10⁻⁵.

   Learn more about charges here:

          brainly.com/question/26092261

                    # SPJ4

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Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave
Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

6 0
3 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
3 years ago
A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magne
ser-zykov [4K]

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R =  16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

8 0
3 years ago
The whooping crane (Grus americana) is the tallest bird native to North America. It almost went extinct in the 1900s; in 1938, t
grin007 [14]

Answer:

The value  is   a =  2.7183

Explanation:

From the question we are told that

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  The  exponential relationship is  N  =  na^{kt}

Now from the given equation we have that

       N  =   e^{2.85 + 0.039t}

       N  =  17.29 e^{0.039t}

So comparing this equation obtained an the given  exponential relationship we have that

      n = 17.29

       k  =  0.039

        a =  e =  2.7183

        a =  2.7183

 

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3 years ago
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Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic radiation. Electromagnetic radiation can be described in terms of a stream of mass-less particles, ...

The electromagnetic spectrum is a map of all the types of light that we can identify. It separates all the types of light by wavelength because that directly relates to how energetic the wave is. More energetic wave

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4 0
3 years ago
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