All categories

3 years ago

A motorist travels a distance of 406 km during a 7 hr period. What was the average speed in (a) km/hr and (b) m/sec?

Physics

2 answers:

Jobisdone [24]3 years ago

8 0

Answer:

The average speed was 58 $\frac{km}{hr}$ or 16.11 $\frac{m}{sec}$

Explanation:

Speed is a physical quantity that expresses the relationship between the space traveled by an object, the time used for it and its direction. Its unit of measurement in the International System (S.I.) is the meter per second $\frac{m}{s}$ .

So:

$speed=\frac{distance}{time}$

In this case you know:

distante= 506 km
time=7 hr

Replacing:

$speed=\frac{406 km}{7 hr}$

Solving:

$speed=58 \frac{km}{hr}$

Taking into account that 1 km is 1000 m and 1 hour is 60 minutes, which in turn is 60 seconds, the unit conversion is performed:

$speed= 58\frac{km}{hr} * 1 000\frac{m}{km} *\frac{1}{60} \frac{hr}{minutes} *\frac{1}{60} \frac{minute}{sec}$

$speed=16.11 \frac{m}{sec}$

The average speed was 58 $\frac{km}{hr}$ or 16.11 $\frac{m}{sec}$

Sholpan [36]3 years ago

4 0

The motorist travels (a) 58 km/h and (b) ~16.1 m/sec

You might be interested in

How much force to move 18 kg at a rate of 3m/s

Stolb23 [73]

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

4 0

3 years ago

How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?

Marrrta [24]

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by $V = R I$

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

$115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes$

8 0

3 years ago

A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse

bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, $f=4.2\times 10^{14}\ Hz$

Time, $t=3.2\times 10^{-11}\ s$

(a) Let $\lambda$ is the wavelength of this light. It can be calculated as :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}$

$\lambda=7.14\times 10^{-7}\ m$

$\lambda=714\ nm$

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

$n=f\times t$

$n=4.2\times 10^{14}\times 3.2\times 10^{-11}$

n = 13440

Hence, this is the required solution.

8 0

2 years ago

If the potential difference across a 12-ohm resistor is 6 volts, the current through the resistor is?

larisa86 [58]

Current is inversely proportional to the resistance of the resistor and directly to the potential difference across it.

I = V/R = 6/12 = 0.5 A

4 0

3 years ago

For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

3 0

3 years ago