Ask question

Ask question

All categories

3 years ago

8

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v

it+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position = 5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

<u>a = 1.72 m/s²</u>

You might be interested in

Why do electric fish need to force electric charges to move?

Anastasy [175]

The electric eel generates large electric currents by way of a highly specialized nervous system that has the capacity to synchronize the activity of disc-shaped, electricity-producing cells packed into a specialized electric organ. Hope this helps!! Good luck

3 0

3 years ago

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago

Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou

umka21 [38]

Ans: Beat frequency = $f_b$ = 4Hz

Explanation:
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = $f_b = |f_2 - f_1|$
<span>
Since,
</span> $f_1 = 480Hz$
$f_2 = 484Hz$

Therefore,
Beat frequency = $f_b = |484 - 480|$

=> Beat frequency = $f_b = 4Hz$
-i

7 0

3 years ago

Your performance on the pacer is used to measure which components of health- related fitness?

77julia77 [94]

Your performance on the pacer test is used to measure cardiovascular components of health-related fitness

8 0

4 years ago

An ice rink is 200 feet long and 85 feet wide. The four corners of the rink are rounded to a radius of 24 feet. What is the surf

Nastasia [14]

Answer:

16,506 ft²

Explanation:

There are different ways you can divide the area using rectangles and circles. One way is to find the area of the entire width and length, then subtract the empty areas in the corners.

If we take the empty areas and put them together, we find their area is the area of a square minus the area of a circle.

A = (2r)² − πr²

A = 4r² − πr²

A = (4 − π) r²

So the area of the rink is:

A = WL − (4 − π) r²

A = (85)(200) − (4 − π) (24)²

A ≈ 16,506 ft²

5 0

3 years ago