Which one is correct?

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

Alex, parked by the side of an east-west road, is watching car P, which is moving in a westerly direction. Barbara, driving east

Alexxandr [17]

Answer:

$v_{PB} = 130\ km/h$

Explanation:

Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:

$Absolute\ Velocity\ of\ Car\ P = v_{P} = -78\ km/h$

And the absolute velocity of Barbara's Car is given as: $Absolute\ Velocity\ of\ Barbara's\ Car = v_{B} = 52\ km/h$

Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:

$Velocity\ of\ Car\ P\ measured\ by\ Barbara = v_{PB} = v_{B}-v_{P}\\\\v_{PB} = 52\ km/h-(-78\ km/h)$

$v_{PB} = 130\ km/h$

6 0

3 years ago

An apparatus like the one Cavendish used to ﬁnd G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.

Ber [7]

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ .

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$

So, the magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ . Hence, this is the required solution.

5 0

3 years ago

Find the kinetic energy of a ball of mass 200 grams moving at a speed of 20 m/s

Aleks04 [339]

Answer:

40 J

Explanation:

$KE = \frac{1}{2} mv^{2} \\KE = \frac{1}{2} (0.200 kg)(20 m/s)^{2} \\KE = 40 J$

6 0

3 years ago

Read 2 more answers

Which statement best describes what Kendall can do?

Vesna [10]

Answer: Explanation:

She can measure the volume and mass of the marble, the volume and mass of the water, and the volume and mass of the graduated cylinder. She can measure the volume and mass of the marble, the volume and mass of the water, and the mass of the graduated cylinder.

5 0

3 years ago