I am in love guess who?

A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a

Rama09 [41]

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

$V(t) \ = \ V_0 \ + \ a \ t$

where $V_0$ is the initial speed, and a its the acceleration.

<h3>First half of the problem</h3>

Starting at rest, the initial speed will be zero, so

$V_0 = 0$

the final speed is

$V(t_{f1}) = 23 \frac{m}{s}$

and the acceleration is

$a = 2.3 \frac{m}{s^2}$ .

Taking all this together, we got

$V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}$

$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

$10 s = t_{f1}$

So, for the first half of the problem we got a time of 10 seconds.

<h3>Second half of the problem</h3>

Now, the initial speed will be

$V_0 = 23 \frac{m}{s}$ ,

the acceleration

$a=-1.0 \frac{m}{s^2}$ ,

with a minus sign cause its slowing down, the final speed will be

$V(t_{f2}) = 0$

Taking all together:

$V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}$

$23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}$

$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

So, for the first half of the problem we got a time of 23 seconds.

<h3>Total time</h3>

$t_total = t_{f1} + t_{f2} = 33 \ s$

5 0

3 years ago

A car accelerates uniformly from rest and reaches a speed of 22.5 m/s in 8.96 s. Assume the diameter of a tire is 58.9 cm. (a) F

Murljashka [212]

Answer:

54.4747649021

38.2003395586 rad/s

Explanation:

Acceleration

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{22.5-0}{8.96}\\\Rightarrow a=2.51116071429\ m/s^2$

Distance covered

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 2.51116071429\times 8.96^2\\\Rightarrow s=100.8\ m$

Number of revolutions

$n=\dfrac{s}{\pi d}\\\Rightarrow n=\dfrac{100.8}{\pi 0.589}\\\Rightarrow n=54.4747649021$

Number of revolutions is 54.4747649021

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{22.5}{0.589}\\\Rightarrow \omega=38.2003395586\ rad/s$

The final angular speed is 38.2003395586 rad/s

8 0

4 years ago

When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde

Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

$F=\dfrac{kq_1q_2}{r^2}$

r is the separation between charges

$F\propto \dfrac{1}{r^2}$

$r=\sqrt{\dfrac{1}{F}}$

If F'= 2F

$r'=\dfrac{1}{\sqrt{2F} }$

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. $r'=\dfrac{1}{\sqrt{2F} }$ . Hence, this is the required solution.

6 0

3 years ago

Help!! It's has multiple answers

Sophie [7]

-- below the horizon
-- clear sky

7 0

3 years ago

What is the dimensional formula of thermal conductivity?

Inga [223]

Answer:

The dimension of thermal conductivity is M1L1T−3Θ−1, expressed in terms of the dimensions mass (M), length (L), time (T), and temperature (Θ). Other units which are closely related to the thermal conductivity are in common use in the construction and textile industries.

Explanation:looked it up for ya

6 0

3 years ago