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2 years ago

Calculate the total resistance in the circuit

Physics

2 answers:

vazorg [7]2 years ago

4 0

Answer:

The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.

SOVA2 [1]2 years ago

4 0

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.

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A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

How can i prove the conservation of mechanical energy?

FinnZ [79.3K]

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

5 0

2 years ago

STATE THE LAWS OF CONSERVATION OF MOMENTUM

dalvyx [7]

$\Huge \mid \underline {\mathcal {{{\color{purple}{Answer...}}}}} \mid$

When two bodies collide with each other in the absence of an external force, then the total final momentum of the bodies is equal to their total initial momentum.

6 0

3 years ago

In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

Sphinxa [80]

Answer: $\frac{V^{2}}{r}$

Explanation:

According to Newton's 2nd Law of motion the force $F$ is proportional to the mass $Fm$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, the equation for the Centripetal force is:

$F=\frac{mV^{2}}{r}$ (2)

Where:

$V$ is the velocity

$r$ is the radius of the circular motion

Making (1) and (2) equal:

$m.a=\frac{mV^{2}}{r}$ (3)

Hence:

$a=\frac{V^{2}}{r}$ This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0

3 years ago

Read 2 more answers

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$

$\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N$

$R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N$

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0

3 years ago

Calculate the total resistance in the circuit​

Calculate the total resistance in the circuit