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nikdorinn [45]
3 years ago
5

A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?

Physics
1 answer:
TiliK225 [7]3 years ago
7 0

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

           Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

                            R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}

                            =\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}

                            =\sqrt{25+4+0} m  [cos90= 0]

                            =\sqrt{29} m

                            = 5.38516 m   [ans]

                     


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Explanation:

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Two resistors A and B are arranged in series in one branch of a parallel arrangement. The other branch contains a single resisto
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Answer:

Explanation:

A and B are in series , Total resistance = Ra + Rb

This resistance is in parallel with single resistor C

Equivalent resistance Re = Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )]

Now this combination is in series in single resistance D .

Total resistance = Rd + Re

= Rd + { Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )] }

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3 years ago
At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does
insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s is the time constant of the circuit

At time t=0,

V_L(0) = V e^0 = V = 20 V

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

V_R = V-V_L = 20 V-20 V=0

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

V_R = V-V_L = 20 V- 0 = 20 V

(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V

So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

We said that the two voltages will be equal when

V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

I_0 = 3.20 A

The problem says this current is stable: this means that we are in a situation in which t>>\tau, so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0
3 years ago
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

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3 0
3 years ago
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