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den301095 [7]
3 years ago
14

Two metal spheres are suspended from strings. The

Physics
1 answer:
IRINA_888 [86]3 years ago
6 0

Answer:

The answer to your question is the letter B. A repulsive force of 8.0 x 10³N

Explanation:

Data

Charge 1 = -2.0 x 10² C

Charge 2 = -4.0 x 10⁻⁸ C

K (Coulomb constant) = 9 x 10⁹ Nm²/C²

distance = 3 m

Formula

F = K (q₁ x q₂) / r²

Substitution

F = 9 x 10⁹ (-2.0 x 10²)(-4.0 x 10⁻⁸) / 3²

Simplification

F = 9 x 10⁹ (0.000008) / 9

F = 72000 / 9

Result

F = 8000 C = 8 x 10³ N

Conclusion

As both spheres are negative the force is negative with magnitude 8x 10³N

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A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,
masha68 [24]

Answer:

a) The car was moving at a speed of 29.167\ m.s^{-1}

b) The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c) f_o=1283.33\ Hz

Explanation:

Given:

  • original frequency of the source, f=1200\ Hz
  • speed of the source, v_s=0\ m.s^{-1}
  • velocity of the obstacle car be, v_o
  • speed of sound, s=350\ m.s^{-1}
  • observed frequency, f_o=1100\ Hz

<u>Using the equation from the Doppler's effect:</u>

\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}

\frac{1100}{1200} =\frac{(350+v_o)}{350-0}

v_o=-29.167\ m.s^{-1}

a)

The car was moving at a speed of 29.167\ m.s^{-1}

b)

The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, v_s=50\ m.s^{-1}

Then, f_o=?

Using the Doppler's eq.:

\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}

f_o=1283.33\ Hz

3 0
3 years ago
A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
adelina 88 [10]

Answer:

x = 11.23  m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

    θ = 155 rev (2pi rad / rev) = 310π rad

    α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

    w² = w₀² + 2 α θ  

    w =√ 2 α θ

    w = √(2 4pi 310pi)

    w = 156.45  rad / s

The relationship between angular and linear velocity

    v = w r

    v = 156.45  0.175

    v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

    y = v_{oy} t - ½ g t²

As it leaves the highest point its speed is horizontal

   y = 0 - ½ g t²

   t = √ (-2y / g)

   t = √ (-2 (-0.820) /9.8)

   t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

   x = vox t

   x = 27.38 0.41

   x = 11.23  m

5 0
4 years ago
A 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1. The contact time is 0.01 seconds. What
Ilia_Sergeevich [38]

Answer:

1.5 kgms⁻¹

Explanation:

Momentum can be defined as "<em>mass in motion</em>."  

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motiom 

when there is a change in the velocity , it creates a change in momentum also

when we consider that we can mathematically show this,In terms of an equation,

Change in momentum    (ΔΡ) = m(Δv)

where (Δv) - change in velocity

<em>(Δv) = final velocity - initial velocity</em>

Change in momentum (ΔΡ) = m(Δv)

                                             = 0.1×([55-40])

                                             = 1.5 kgms⁻¹

     

7 0
3 years ago
an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20
Schach [20]
Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J
5 0
3 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
madreJ [45]

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

8 0
3 years ago
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