A rock is dropped from a bridge. What happens

What is the average acceleration of a tennis ball that has an initial velocity of 6.0 m/s [E] and a final velocity of 7.3 m/s [W

Marizza181 [45]

Given :

The average acceleration of a tennis ball that has an initial velocity of 6.0 m/s.

and a final velocity of 7.3 m/s.

It is in contact with a tennis racket for 0.094 s

To Find :

The average acceleration of the tennis ball.

Solution :

We know, average acceleration is given by :

$a_{avg}=\dfrac{Final \ velocity-Initial\ velocity}{Time\ Taken}\\\\a_{avg}=\dfrac{7.3-6.0}{0.094}\ m/s^2\\\\a_{avg}=13.83\ m/s^2$

Therefore, average velocity is given by 13.83 m/s².

Hence, this is the required solution.

7 0

2 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

7 0

2 years ago

A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is

neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed $v_{r}$ =2 m/s

velocity of motorboat relative to water is $v_{m/r}$ =4.2 m/s

so speed will be:

a) $v_{m}$ = $v_{r}$ + $v_{m/r}$

solving graphically

$v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}$

=4.7 m/s

Ф= $tan^{-1} (\frac{v_{r}}{v_{m/r}} )$

=25.5°(south of east)

b) time to cross the river: t= $\frac{w}{v_{m/r}}$ = $\frac{500}{4.2}$ =119 s

c) d= $v_{r}t$ =(2)(119)=238 m

note :

pic is attached

6 0

3 years ago

A 20 N force is applied to an object causing it to move 10 m. How much work was done on the object? How much energy was needed t

olga2289 [7]

Answer:

Here, force=20N and displacement=10m

Work=Force×Displacement=20N×10m=200Nm

3 0

3 years ago

In the Bronsted-Lowry Theory of acids and bases, an acid is this.

madreJ [45]

Answer: An acid is a substance that donates a proton and produces a conjugate base.

Explanation:

According to Bronsted-Lowry theory, an acid is a substance that donates a proton and produces a conjugate base while a base is a molecule or ion which accepts the proton.

An example of Bronsted-Lowry acid and base is Ethanoic acid, CH3COOH and hydroxide ion, OH- respectively as shown in the reaction below

CH3COOH(aq) + OH-(aq) <---> CH3COO-(aq) + H2O(l)

Thus, ethanoic acid acts as an acid by donating a proton to the hydroxide ion which accepts it, thus producing ethanoate ion, CH3COO- as a conjugate base.

6 0

3 years ago