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Levart [38]
3 years ago
7

Pls answer ASAP pls and thnk you my friends

Physics
1 answer:
rodikova [14]3 years ago
6 0

Answer:

Since this is old, hope you don't mind if I scoop up these points

Explanation:

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Three identical train cars, coupled together are rolling east at 2.0 m/s. A fourth car traveling east at 4.0 m/s catches up with
Ann [662]

Answer:

The value is  v  =  2 \ m/s

Explanation:

From the question we are told that

   The velocity of the each of the three cars is u_1 = u_2 = u_3 =  2 \  m/s

    The velocity of the fourth car is  u_4 =  4 \ m/s

    The initial velocity of the fifth car u_5 =  0 \ m/s

Generally from the law of momentum conservation we have that

    m_1 u_1 + m_2 u_2 + m_3 u_3 +m_4u_4 + m_5u_5 =  [m_1   + m_2 + m_3 +m_4+ m_5]v

Given that the cars are identical then their mass will be the same

i.e

    m_1 =m_2 = m_3 = m_4 = m_5 =  m

=>   [u_1 + u_2 +  u_3 +u_4 + u_5]m =  5mv

=>   2+ 2 +  2 +4 + 0 =  5v

= >   v  =  2 \ m/s

8 0
3 years ago
: Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in
Aneli [31]

Answer:

a) 096V  b) 0.0288A  c) 0.3456W

Explanation:

a) Vp/Vs= Np/Ns

 120/Vs= 500/4

Vs= 096V

b) Np/Ns= Is/Ip

500/4= 3.6/Ip

Ip= 0.0288A

c) P= VI

P=(120)(0.0288)

P= 0.3456W

8 0
3 years ago
A series RCL circuit consists of a 50 Ω resistor, 500 Ω capacitor, and a 500 Ω inductor. What is the angle of impedance?
lbvjy [14]

Answer:

Zero

Explanation:

here, the inductive reactance and the capacitive reactance is same, so this is the condition for resonance.

In the condition for resonance,

the circuit and the voltage in the circuit is in the same phase and the impedance of the circuit is minimum which is equal to the resistance of the circuit.

The phase angle is given by

tan\phi =\frac{X_{L}-X_{C}}{R}

Ф = 0

5 0
3 years ago
Which statement about force is incorrect
garik1379 [7]

Answer:

What are the options?

Explanation:

4 0
3 years ago
A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang
Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]

E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]

E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]

Since \lambda = \frac{Q}{L}

where;

L = length

Q = charge

λ =  density of the charge;

then substituting \frac{Q}{L} for λ, we have :

E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]

E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}

substituting our given parameter; we have:

E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0
3 years ago
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