The velocity of a car increases from 10 km/h to 50 km/h in 5 seconds. What is its acceleration?

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

Please help me with this (with explanation)

Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of t hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (t - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

x = (22.0 km/hr) • ((t - 1/3) hr) ≈ (22.0 km/hr) t - 7.33 km

If the cyclist's average speed over the total time t was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = x / t

Replace x with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) t - 7.33 km) / t

Solve for t :

17.5 km/hr = 22.0 km/hr - (7.33 km) / t

(7.33 km) / t = 4.5 km/hr

t = (7.33 km) / (4.5 km/hr)

t ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

x ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = x / t

you can instead solve for t in terms of x, then plug that into the distance equation.

t = x / (17.5 km/hr)

then

x ≈ (22.0 km/hr) (x / (17.5 km/hr)) - 7.33 km

x ≈ 1.25714 x - 7.33 km

0.25714x ≈ 7.33 km

x = (7.33 km) / 0.25714 ≈ 28.5 km

6 0

2 years ago

List and describe the top 3 qualities of an effective leader?

s2008m [1.1K]

BEING HONEST , OPTIMISTIC,DISCIPLINED, PUNCTUAL ,TRUTHFUL AND GOOD ORATOR(SPEAKER) , AND HOPEFUL ARE OTHER BASIC QUALITIES AN EFFECTIVE LEADER MUST POSSESS. ALSO,HE MUST NEVER DIFFERENCE BETWEEN ANY OF HIS FOLLOWERS OR DO PARTIALLY AT ANY TIME . HE SHOULD TREAT EVERY ONE EQUALLY.

Explanation:

7 0

3 years ago

I need to know which is represented as a element

Tema [17]

Answer:

the yellow one

Explanation:

2 of the same elements resolute as the same element

6 0

3 years ago

Mars completes one orbit around the Sun in approximately two Earth years. Mars orbits at an average distance to the Sun of about

satela [25.4K]

Answer:

about 0.8 times

Explanation: MARS is a planet named after a Roman God of War, it is the fourth planet from the sun and the Second smallest planet following Mercury.

The orbital Speed of a planet is the speed with which the planet uses to completely go round the Sun.

The orbital Speed of the Earth is 29.78kilmeters per Second.

The orbital Speed of Mar is 0.8*29.78kilometers per second/1

=23.824kilmeters per second.

3 0

3 years ago