9. What do people playing pool use to determine their shots?

Calculate the acceleration of a 1400-kg car that stops from 39 km/h "on a dime" (on a distance of 1.7 cm).

Vinil7 [7]

Answer:

$a=-3449.67\frac{m}{s^2}$

Explanation:

The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:

$v_f^2=v_0^2+2ax$

We convert the initial speed to $\frac{m}{s}$

$39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}$

The car stops, so its final speed is zero. Solving for a:

$a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}$

4 0

3 years ago

A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes

Stolb23 [73]

Answer:

a) -4 N

b) +4 N

Explanation:

Draw a free body diagram for each block.

For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.

For the small block, there is 1 force, F pushing to the right.

There are also weight and normal forces in the vertical direction, but we can ignore those.

Sum of forces on the large block in the x direction:

∑F = ma

12 − F = 4a

Sum of forces on the small block in the x direction:

∑F = ma

F = 2a

2F = 4a

Substitute:

12 − F = 2F

12 = 3F

F = 4

The small block pushes on the large block 4 N to the left (-4 N).

The large block pushes on the small block 4 N to the right (+4 N).

4 0

3 years ago

Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the

Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

Observer A is stationary and Observer B is moving.
Observer A and Observer B are moving at different speeds relative to each other.

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is due to the motion of the observer or the source of the frequency (or both).

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

Observer A is stationary and Observer B is moving.
Observer A and Observer B are moving at different speeds relative to each other.

If you want to learn more, you can read:

brainly.com/question/17107808

6 0

2 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$