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skelet666 [1.2K]
4 years ago
7

A 0.7 kg lab cart moving to the right at 0.15 m/s collides with a 0.5 kg lab cart moving to the right at 0.10 m/s. After the col

lision, the 0.7 kg cart is moving to the right at 0.08 m/s. Calculate the velocity of the 0.5 kg lab cart after the collision.
Physics
1 answer:
Jet001 [13]4 years ago
6 0

Answer:

0.198 m/s

Explanation:

m1 = 0.7 kg, u1 = 0.15 m/s,

v1 = 0.08 m/s

m2 = 0.5 kg, u2 = 0.1 m/s

Let the speed of 0.5 kg is v2 after the collision.

By using the conservation of momentum

Momentum before collision = momentum after collision

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.7 x 0.15 + 0.5 x 0.1

= 0.7 × 0.08 + 0.5 × v2

0.099 = 0.5 v2

v2 = 0.198 m/s

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Q.3. The equivalent resistance across AB is:<br> (a)1<br> (c)2<br> (b)3<br> (d)4
sp2606 [1]

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

7 0
3 years ago
How to solve for the coefficient of friction
astraxan [27]
Their are two coefficients of friction, static and kinetic, regardless, they have basically the same formula:

u*N = F... aka
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3 years ago
Define nortons theorem​
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Answer:

<em><u>In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.</u></em>

8 0
3 years ago
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Soloha48 [4]

Answer:

8) d

9) c

<em>Hope this helps! :)</em>


5 0
3 years ago
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our
Gnom [1K]

Answer:

 f = v / 4L

the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Explanation:

In wind instruments the wave speed must meet

          v = λ f

          λ = v / f

from v is the speed of sound that depends on the temperature

          v = v₀ \sqrt{1+ \frac{T [C]}{273} }

where I saw the speed of sound at 0ºC v₀ = 331 m/s  the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation

            (F -32) 5/9 = C

            76ºF = 24.4ºC

            45ºF = 7.2ºC

           

With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside

at T₁ = 24ºC               v₁ = 342.9 m / s

at T₂ = 7ºC                 v₂ = 339.7 m / s

To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change

              λ / 4 = L

              λ= 4L

              v / f = 4L

              f = v / 4L

Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

3 0
3 years ago
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