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erma4kov [3.2K]

2 years ago

13

Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car w

ith an acceleration of 1.80 m/s2. How long does it take car A to overtake car B

1 answer:

goblinko [34]2 years ago

5 0

Answer:

car A reaches and immediately overtakes the car B at 22.56 s.

Explanation:

After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:

$x_{A}=300+x_{B}$

Now, we can rewrite this equation in terms of speed and time

$V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t$

Where:

V(iA) is the initial speed of car A

V(iB) is the initial speed of car B

t is the time when car A reaches the car B

a is the acceleration

$18t+\frac{1}{2}1.8t^{2}=300+25t$

$0.9t^{2}-7t-300=0$

Solving this quadratic equation for t, and taking just the positive value, we will have:

t=22.56 s

Therefore, car A reaches and immediately overtakes the car B at 22.56 s.

I hope it helps you!

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Please select the word from the list that best fits the definition

enyata [817]

Answer:

conductor

Does not easily transfer electricity

8 0

2 years ago

Lindsay is planning a flight from St. Catharines to Hamilton, which lies due west of St. Catharines. Her aircraft flies at a spe

olga nikolaevna [1]

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.

From this question, the plane is still up in the air.

We have wind blowing in [W 60° N ]

To solve the problem we have to make use of the sine rule

$\frac{SinA}{a}=\frac{SinB}{b} =\frac{SinC}{c}$

We put the values in the equation, we have:

50/Sinθ = 200/sin60°

The next step is to cross multiply

50 x sin60° = 200Sinθ

50 x 0.8660 = 200sinθ

We make Sin θ the subject

Sine θ = 43.30/200

sine θ = 0.2165

we find the value of θ

θ = sine⁻¹(0.2165)

θ = 12.50

So Lindsay has to fly this plane towards this direction

[W 12.5° S]

Here is a similar question brainly.com/question/13338067?referrer=searchResults

7 0

2 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

<em>Current flowing in the wire, I = 4.00 mA</em>
<em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
<em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
<em>Length of wire, L = 2.00 m</em>
<em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

3 0

3 years ago