Ask question

Ask question

All categories

3 years ago

14

ou are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You pl

an to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40.0 km/h for 43.0 km. What would be the least speed (in km/h) needed for the rest of the trip to arrive in time for the interview?

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

133.62 kmh.

Explanation:

Time provided = 3.25 hours.

Distance to be covered 300 km

Times spent in first 100 km = 1 hour

Time spent in next 43 km

= 43 / 40 = 1.075 hours

Total time spent = 2.075 hours

Total distance covered = 143 km

Distance remaining = 300 - 143

=157 km .

Time remaining = 3.25 - 2.075

= 1.175

Speed required = Distance remaining / time remaining

= 157 / 1.175

= 133.62 kmh.

You might be interested in

A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m\s. The bullet is brought to rest in a

prohojiy [21]

Answer:

1) F = 24 N

2) Distance = 1 m

Explanation:

We are given;

Mass; m = 120 g = 0.12 kg

Initial velocity; u = 20 m/s

Final velocity; v = 0 m/s since it came to rest.

Time; t = 0.1 s

We can calculate acceleration from Newton's first equation of motion;

a = (v - u)/t

a = (0 - 20)/0.1

a = -200 m/s²

1) magnitude of the resistance will be;

F = ma

F = 0.12 × (-200)

F = -24 N

Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N

2) To find the distance moved by the bullet, we know that;

Distance = Average speed × time

Thus;

Distance = ((v + u)/2) × t

Distance = ((0 + 20)/2) × 0.1

Distance = 1 m

4 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi

USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}\\$ .

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\$ .

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$ .

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$

Now to determine the extension,

$L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\$

4 0

3 years ago

The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val

yaroslaw [1]

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

4 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

4 years ago

Some populations have cooperative relationships. This is where:

Ilia_Sergeevich [38]

C. Members of the same species work together for survival

5 0

3 years ago