Ask question

Ask question

All categories

3 years ago

15

You riding in a car traveling at a speed of 73 mi/hr in a. 60 mi/hr zone, when suddenly you spot a police car parked 20 ft ahead

of you. Do you hit the brakes and slow down with an acceleration of -2 ft/sec,for a total time of 3 seconds. Do you get a ticket?

2 answers:

fenix001 [56]3 years ago

5 0

I think yes i do.. cuz using yr information i get the distance i travelled before stopping become 49.62144 m, using the formula distance travelled/time taken= (final velocity+initial velocity)/2.
I am not sure tht its the right answer but using my calculation I got this so hope there! :)

Nataliya [291]3 years ago

4 0

Answer:

You get a ticket you speed is around 68 mi/hr

Explanation:

$V_{f} =V_{i} + a*t$

$V_{i} = 73 \frac{mi}{hr} *\frac{1 hr}{3600 s} * \frac{1609,34m}{1 mi} = 32.6339 \frac{m}{s}$ , $a=-2 \frac{ft}{s^{2} } * \frac{0.3048m}{1 ft} =-0.6096 \frac{m}{s^{2} }$

$V_{f} =32.6339\frac{m}{s} -0.6096 \frac{m}{s^{2} } *3 s$

$V_{f} = 30.8051\frac{m}{s} *\frac{3600 s}{hr} * \frac{1 mi}{1609.34m} = 68.909 \frac{mi}{hr}$

You still in trouble because the speed is still above the allowed

You might be interested in

Help with this please.

gregori [183]

Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>

8 0

3 years ago

States that there is an exchange of materials when two objects come into contact with each other

pishuonlain [190]

"Edmond Locard" states that there is an exchange of materials when two objects come into contact with each other.

<u>Explanation:</u>

A French criminologist who was popular as the "Sherlock Holmes of France," the pioneer in forensic science named as Dr. Edmond Locard. He articulated forensic science's fundamental principle "Each touch leaves a trace." This became known as Locard's philosophy of exchange. A Locard hypothesized that each and every time you touch another person, place or object, the result would be an exchange of materials. Burglars, for instance, will leave evidence of their existence behind and take traces with them too.

3 0

3 years ago

Object a has a mass of 20 g

Anarel [89]

And.. where is the rest of the question?

7 0

3 years ago

A block of mass, m, sits on the ground. A student pulls up on

kakasveta [241]

Answer a

Explanation: a

3 0

3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

3 years ago