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4 years ago

12

After the driver first notices the obstacle, the car moves uniformly for a time interval t1−t0=t before the brakes are applied.

Find the distance x1−x0 covered by the car in the time interval before the brakes are applied and after the driver sees the obstacle.

1 answer:

loris [4]4 years ago

3 0

Answer:

V(t1-t0)

Explanation:

Moving 'uniformly' means constant velocity (speed). the formula for constant speed motion is $V = \frac{distance}{time}$ =( change in position/ change in time)

where,

V is speed

given in the statement :

change in time = t = t1-t0

let the constant speed be ' V '

disance = X = X1-X0

applying the above mentioned formula: V = $\frac{X}{t}$

V = X/t

X = Vt

the distance X1-X0 = Vt =V(t1-t0)

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What happens to a circuit's resistance (R), voltage (V), and current (1) when

hammer [34]

Answer:

D.

R increases

V is constant

I decreases

Explanation:

The resistance of a wire is given by the following formula:

$R = \frac{(Resistivity)(L)}{A}$

It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, <u>the resistance of circuit increases</u>.

The <u>voltage in the circuit will be constant</u> as the voltage source remains same and it is not changed.

Now, we can use Ohm Law:

V = IR

at constant V:

I ∝ 1/R

it means that current is inversely proportional to resistance. Hence, the increase of resistance causes <u>the current in circuit to decrease.</u>

Therefore, the correct option will be:

<u>D.</u>

<u>R increases </u>

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3 years ago

You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

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<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

<u>Now the force on the given charge can be given as:</u>

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$F=150\times 4\times 10^{-9}$

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4 years ago

The value of 1.0004 to the power 1 by 2 using Binomial approximation is

IgorLugansk [536]

Given:

The given value is $(1.0004)^{\frac{1}{2}}$ .

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

$(1.0004)^{\frac{1}{2}}$

It can be written as:

$(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}$

$(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004$ $[\because (1+x)^n=1+nx]$

$(1.0004)^{\frac{1}{2}}=1+0.0002$

$(1.0004)^{\frac{1}{2}}=1.0002$

Therefore, the approximate value of the given expression is 1.0002.

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3 years ago

On a balanced seesaw, a boy three times as heavy as his partner sits

slega [8]

Answer:

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If we substitute this into the equation, we find:

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and by simplifying:

$3 d_1 = d_2\\d_1 = \frac{1}{3}d_2$

which means that the boy sits at 1/3 the distance from the fulcrum.

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This problems a perfect application for this acceleration formula:

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