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3 years ago

12

A vehicle is moving at a speed of 20m/s suddenly the driver sees a fallen tree on the road, he instantly applies the brakes that

subject the car to a deceleration of -2m/s².Find
a) Time required by the car to stop.
b) distance covered before stopping.

1 answer:

NeTakaya3 years ago

6 0

Answer:

b) Distance covered before stopping.

Explanation:

If you want to find the distance d required to come to a stop starting at some initial speed v, with braking acceleration a, use the kinematic relation

vf2 - vi2 = 2ad

vi = initial speed at the moment braking begins = v

vf = final speed = 0 (comes to a full stop)

-v2 = 2ad

d = -v2/(2a)

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Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

Learn more about current and voltage:

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4 0

3 years ago

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 ) x 25 = 2083.33 mV = 2.083 V.

7 0

3 years ago

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th

Marina CMI [18]

Answer:

The radius of the curve is 9,183.67 m.

Explanation:

Given;

velocity of the jet plane, v = 600 m/s

acceleration of the jet plane, a = 4g = 4 x 9.8 m/s² = 39.2 m/s²

The radius of the curve is calculated from centripetal acceleration formula as given below;

$a = \frac{v^2}{r} \\\\r = \frac{v^2}{a} \\\\r = \frac{600^2}{39.2} \\\\r = 9,183.67 \ m$

Therefore, the radius of the curve is 9,183.67 m.

3 0

3 years ago

Please don’t troll. I need someone to actually answer these questions.

mash [69]

Answer:

-3+3 i think this is the answer

Explanation:

i think you can ask someone else sorry

4 0

3 years ago

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the

Illusion [34]

angular vel to tangential vel

v=r omega

v = 56 x 100/60 x 2 pi

v = 56x5/3x6

v=560m/s as estimate

100 revs, 5.00m

5 0

3 years ago

Read 2 more answers