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4 years ago

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You're driving down the highway late one night at 18 m/s when a deer steps onto the road 35 m in front of you. your reaction tim

e before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 11 m/s

1 answer:

DedPeter [7]4 years ago

8 0

You are not asking a clear question

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UPVOTES FOR EVERY ANSWER!!!!!!!!

Goshia [24]

<span>A. The magnetic force increases.

F is inversely proportional to r</span>²<span>

Hope this helps!</span>

7 0

3 years ago

Read 2 more answers

If an investigator wanted to reproduce the results with a beam of protons, what would require to get a strong m = 1 reflection a

Elis [28]

When the investigator wanted to reproduce the results with a beam of protons, the thing required is A. Lower speed particles than for the electron beam.

<h3>What is proton?</h3>

It should be noted that a proton simply means a stable subatomic particle that has a positive electric charge.

In this case, when the investigator wanted to reproduce the results with a beam of protons, the thing required is a lower speed particles than for the electron beam.

Learn more about protons on:

brainly.com/question/1805828

3 0

3 years ago

slader) Two charges are arranged at corners of a square which has a side length of L = 0.25 m. The values of the charges are q1

nataly862011 [7]

Explanation:

Formula to calculate the electric potential is as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

Putting the given values into the above formula as follows.

$V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

= $\frac{9 \times 10^{9}}{0.25}[\frac{-3.3}{\sqrt{2}} + 4.2] \times 10^{-6}$

= $6.72 \times 10^{4} V$

Hence, electric potential at point A is $6.72 \times 10^{4} V$ .

Now, the electric potential at point B is as follows.

$V_{B} = \frac{kq_{1}}{L} + \frac{kq_{2}}{\sqrt{2}L}$

= $\frac{9 \times 10^{9}}{0.25} [-3.3 + \frac{4.2}{\sqrt{2}}] \times 10^{-6}$

= $-1.19 \times 10^{4} V$

Hence, electric potential at point B is $-1.19 \times 10^{4} V$ .

8 0

3 years ago

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

rosijanka [135]

Answer:

A) $\frac{g}{b}(1-e^{-bt})$

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since $\frac{dv}{dt}= g - bv = b( \frac{g}{b} - v)$ ⇒ $\frac{dv}{ \frac{g}{b} - v}= bdt$

So take the integral of both side.

$- ln (\frac{g}{b} - v) = bt + C$

Since for t=0, v = 0 ⇒ $C =- ln (\frac{g}{b})$

$v = \frac{g}{b} + e^{-bt-ln(\frac{g}{b})} = \frac{g}{b}- \frac{g}{b}e^{-bt} = \frac{g}{b}(1-e^{-bt})$

5 0

3 years ago

10)A car is moving from rest and attained a velocity of 80 m/s. Calculate the

Sergio [31]

Answer:

initial velocity (u)=0

final velocity (v)=80m/s

time(t)=5s

acceleration (a)=?

Explanation:

we have

v=u+at

80=0+a×5

5a=80

a=80/5=16m/s²

5 0

3 years ago