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3 years ago

12

A 1000 kg car pushes a 2000 kg car that has a dead battery. When the driver steps on the accelerator, the drive wheels of the ca

r push backward against the ground with a force of 4500 N.

1 answer:

butalik [34]3 years ago

4 0

So what is the question

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The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr

SVETLANKA909090 [29]

Answer:

∅= $63.43^{0}$

Explanation:

z=impedance

$x_{l}$ =2R

R=R

The resultant of the resistances in the circuit is called impedance

$x_{l}$ is inductive reactance of the circuit

R is the resistance of the resistor

z= $\sqrt{xl^{2}+R^2 }$

z= $\sqrt{2R^{2}+R^2 }$

Z= $\sqrt{5R^2}$

Z=R $\sqrt{5}$ ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅= $63.43^{0}$

phase angle is ∅= $63.43^{0}$

3 0

3 years ago

The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :

$\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}$

$t=68.94 m$

8 0

3 years ago

What’s the answer ?????

Mice21 [21]

Answer:

<h2>Refer the attachment for answer and explanation please</h2>

Explanation:

This might surely help you ☺️❤️

7 0

3 years ago

In a series RLC ac circuit, a second resistor is connected in series with the resistor previously in the circuit. As a result of

AnnyKZ [126]

Answer:

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

Explanation:

The impedance of a series circuit is

Z₀² = R² + (X_L-X_C) ²

when we place another resistor in series the initial resistance impedance changes to

Z² = (R + R₂) ² + (X_L - X_C) ²

let's analyze this expression

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

8 0

3 years ago

7.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot

Oksana_A [137]

I believe the answer to this is Zinc

4 0

3 years ago