<img src="https://tex.z-dn.net/?f=If%20%20%5C%3A%20the%20%5C%3A%20%20speed%20%5C%3A%20of%20%5C%3A%20a%20%5C%3A%20car%20%5C%3A%20

All categories

3 years ago

is%20%5C%3A%2020m%20%5C%3A%20%20%7Bs%7D%5E%7B%20-%201%7D%20" id="TexFormula1" title="If \: the \: speed \: of \: a \: car \: is \: 20m \: {s}^{ - 1} " alt="If \: the \: speed \: of \: a \: car \: is \: 20m \: {s}^{ - 1} " align="absmiddle" class="latex-formula">
$then \: what \: is \: its \: speed \: in \: km \: {h}^{ - 1}$
Step by step explanation needed :)
________________________________
Question In a easy way -

If the speed of a car is 20 m s^-1 then what is its speed in km h^-1 ?

Physics

2 answers:

nadezda [96]3 years ago

7 0

Answer:

72 km/h

Explanation:

Speed of car = 20m/s

So, here to find the speed in km/h we will change the metre to km and seconds to hours .

$\frac{20}{1000} \div \frac{1}{3600}$

= 72 km/hr

to change into km divide by 1000

(1 m = 1/1000km)

and,

to change into hours divide by 60*60= 3600

help kiya kya?

Vilka [71]3 years ago

6 0

hope it helps

mark as brainlist

You might be interested in

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A simple pendulum consists of a mass m (the "bob") hanging on the end of a thin string of length l and negligible mass. The bob

svetlana [45]

Answer:

The pendulum frequency is (c) the same, or very close to it

Explanation:

The simple pendulum corresponds to a simple harmonic movement, to reach this approximation in the expression of the force the sine of the angle (θ) approaches an angle value, this is only true for small angles, generally less than 15º

Sine (15th) = 0.2588

The angle in radians is 15º π / 180º = 0.26180.2588 / 0.2618

The difference between these two values is less than 1.2%

for smaller angle the difference is reduced more

Therefore, the period for both the 5º and 10º angles is almost the same

5 0

3 years ago

List 5 possible effects of not adhering to standards of measurement

murzikaleks [220]

Answer:

factual evidence of customer-service levels.

better understanding of cross-functional performance.

enhanced alignment of operations with strategy.

evidence-based determination of process improvement priorities.

detection of performance trends.

better understanding of the capability range of a process.

3 0

3 years ago

The outer planets are similar to the planet Earth; they just happen to be farther away from the Sun.

Harrizon [31]

Answer:

False

Explanation:

The inner planets are called terrestrial planets due to the surfaces are solid (similar to Earth)-made up of heavy metals, either have no moons or few moons.

The outer planets are called Jovian planets or gas giants because they are encased in gas. They all have rings with plenty of moons.

5 0

3 years ago

Read 2 more answers