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3 years ago

9

Two equal length of wire made of the same material but of different diameters have an effective resistance of 0.8 ohm when they

are connected in parallel. If the cross sectional area of one is four times of the other. Calculate the resistance of the thicker wire

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

hfdfg

Explanation:

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A flute plays a note with a frequency of 266 Hz. What is the speed of sound if the wavelength is 1.3 m?

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Wave speed = (frequency) x (wavelength)

= (266 /sec) x (1.3 meters)

= 345.8 meters/sec

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Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed fro

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Answer:

a) Acceleration of runner is 1.33 m/s²

b) Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2$

Acceleration of runner is 1.33 m/s²

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2$

Acceleration of motorcycle is 2.85 m/s²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m$

The runner moves 2.94 m

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m$

The motorcycle moves 84.21 m

The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

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3 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

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Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

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