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3 years ago

Consider a circuit connected in parallel with three resistors, each providing a

Physics

1 answer:

alexgriva [62]3 years ago

3 0

Answer:

I believe that it's c it's a little confusing tho

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How electromagnet works in circuit breaker

Svetradugi [14.3K]

Explanation:

A circuit breaker is an automatic switch that cut off current in a circuit when the current become too large. When the current in a circuit increases, the strength of the electromagnet will increase in accordance; this will pull the soft iron armature towards the electromagnet.

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3 years ago

Right or left? Summer and Winter or day and night?

evablogger [386]

Answer:

The answer is left

Explanation:

Sorry if im worng

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3 years ago

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Which value is equivalent to 7.2 kilograms?

Elis [28]

B is the right answer glad I could help !!

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3 years ago

A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$

$\tau=80\times 0.5\ sin(60)$

$\tau=34.6\ N.m$

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

7 0

3 years ago

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