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3 years ago

Explain all the parts in a complete and closed circuit and how it works.

Physics

1 answer:

Archy [21]3 years ago

6 0

Answer:

In a <u>complete circuit</u>, you have 4 main parts: a power source, a load, connectors and a switch. The power source provides energy for the electricity to travel along the circuits and the load is the device that the circuit is designed to power.

A <u>closed circuit</u> is a circuit where electricity can flow from one end of the battery to the other, all the components of the circuit are conductors. It contains the same parts as a complete circuit.

Hope this helps :)

Explanation:

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never [62]

Answer:

The answer is A

Explanation:

When a rockets thrusters push on the ground the ground pushes back on the rocket with equal force in the opposite direction. Hence the rocket takes off.

Newtons third law of motion states, for every action there is an equal and opposite reaction.

7 0

2 years ago

Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$

6 0

3 years ago

A ball starts from rest. It rolls down a ramp and reaches the ground after 4 seconds. Its final velocity when it reaches the gro

Hunter-Best [27]

If it starts at rest the initial velocity is 0.
For an acceleration, a, and time, t, the velocity is v=at. Since at t=4, v=7, then a=7/4=1.75m/s^2

3 0

3 years ago

Read 2 more answers

A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?

Sergeu [11.5K]

Answer:

216 miles

Explanation:

you times 72 by three

7 0

3 years ago

Read 2 more answers

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago