Ask question

Ask question

All categories

3 years ago

11

Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.

If you were to increase your speed to 60 miles per hour, your stopping distance is now:

2 answers:

Viefleur [7K]3 years ago

8 0

Answer:900 feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

$v^2-u^2=2as$

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph $\approx 88.002 ft/s$

$v^2-u^2=2as$

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

Rus_ich [418]3 years ago

3 0

Answer:

900 feet

Explanation:

Initial Speed, u₁ = 20 mph

Stopping distance, s₁ = 100 feet

Initial Speed, u₂ = 60 mph

Then, the stopping distance can be calculated using the third equation of motion:

$s=\frac{v^2-u^2}{2a}$

There would be same acceleration and final velocity would be zero (v=0).

$s=\frac{0-u^2}{2a}\\ \frac{s_2}{s_1}=\frac{u_2^2}{u_1^2}\\s_2= 100 ft\frac{(60)^2}{(20)^2} =900 feet$

You might be interested in

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

4 0

2 years ago

Read 2 more answers

The graph represents the reaction 2H2 + 02 32H20 as it reaches

Alex

Answer:

C and D

Explanation:

5 0

3 years ago

Read 2 more answers

How is tidal energy obtained

yKpoI14uk [10]

Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion

7 0

3 years ago

What does hydraulic mean?

Harrizon [31]

Answer:

denoting, relating to, or operated by a liquid moving in a confined space under pressure.

4 0

2 years ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago