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Elis [28]
2 years ago
6

The distance between two stations is 180 km. A train takes 2 hours to cover this distance. The speed of the train in m/sec is...

...please help i rlly need it ....Please
Physics
1 answer:
Mamont248 [21]2 years ago
4 0

Answer:

Av = 25 [m/s]

Explanation:

To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

Av=\frac{distance}{time}

where:

Av = speed [km/h] or [m/s]

distance = 180 [km]

time = 2 [hr]

Therefore the speed is equal to:

Av = \frac{180}{2} \\Av = 90 [km/h]

Now we must convert from kilometers per hour to meters per second

90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]

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in the parallel connection the light bulbs shine less than in the series connection

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so the power of each bulb is is

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Can you cool a kitchen by leaving the refrigerator door open while the refrigerator is operating? a. No. Since the refrigerator
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Object A has mass 83.0 g and hangs from an insulated thread. When object B, which has a charge of +140 nC, is held nearby, A is
erastova [34]

Answer:

a) -238 nC  

b) 0.889 N  

Explanation:

Concepts and Principles

<u>Particle in Equilibrium:</u> If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:  

∑F = 0                                                                           (1)  

<u>Coulomb's Law:</u> the magnitude of the electrostatic force exerted by a point charge q1 on a second point charge q2 separated by a distance r is directly proportional to the product of the two charges and is inversely proportional to the square of the distance between them:

F_12 = k*| q1 |*| q2 |/r^2                                                 (2)

where k = 8.99 x 10^9 N  m^2/C^2 is Coulomb constant.  

<u>Given Data  </u>

<em>mA (mass object A) = (83 g)*(1/1000g)=0.09 kg </em>

<em>qB (charge of object B) = (140 nC)*(1/10^9 nC) = 130 x 10^-9 C </em>

<em>Object A is attracted to object B. </em>

<em>Ф(angle made by object A with the vertical) = 7.2°  </em>

<em>(  r (distance between the two objects) = (5 cm) * (1 m/ 100 cm) =0.05 m  </em>

<em>Object A is in equilibrium.  </em>

Required Data

In part (a), we are asked to determine the charge qA of object A.

In part (b), we are asked to determine the tension T in the thread.  

(a) The FBD in Figure 1 shows the forms acting on object A; Fe is the electric force exerted on object A by object B, T is the tension force exerted on the thread, and m_a*g is the gravitational force exerted on object A.  

Model object A as a particle in equilibrium in the horizontal and vertical direction and apply Equation (1) to it:  

∑F_x = F_e-Tsin = 0                                   F_e=Tsin<em>Ф                </em><em>(3)</em>

∑F_y = Tcos<em>Ф - </em>m_a*g= 0                      m_a*g=Tsin<em>Ф                </em><em>(4)</em>

Divide Equation (3) by Equation (4) to eliminate T:

F_e/m_a*g=tan<em>Ф</em>

F_e=m_a*g*tan<em>Ф</em>

Substitute for  F_e by using Coulomb's law from Equation (2):

k*| q_A |*| q_B |/r = m_a*g*tan<em>Ф</em>

Solve for q_A :  

| q_A | = m_a*g*tanФ_r/k*| q_B |

Substitute numerical values from given data:

| q_A | =  238 nC  

Because object A is attracted to object B. it has an opposite negative charge. Therefore, the charge on object A is | q_A | =  -238 nC  

(b)  

Solve Equation (4) for T:  

T = m_a*g/cosФ

Substitute numerical values from given data:

T = (0.09 kg)(9.8 m/s^2) /cos 7.2°  

  = 0.889 N  

4 0
3 years ago
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