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BlackZzzverrR [31]

3 years ago

6

You and your family are traveling to daytona beach for a vacation and your dad wants to get there in 7 hours. If daytona is 725

kilometers away how fast will he have to drive to get there on time

1 answer:

ollegr [7]3 years ago

4 0

Answer:

103.57 Km/h

Explanation:

From the question given above, the following data were obtained:

Distance = 725 Km

Time = 7 hours

Speed =?

Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can calculate how fast he will drive (i.e the speed) in order to get there on time. This is illustrated below:

Distance = 725 Km

Time = 7 hours

Speed =?

Speed = Distance /time

Speed = 725 / 7

Speed = 103.57 Km/h

Thus, to get there on time, he will drive with a speed of 103.57 Km/h

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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

1. The distance between a trough and a crest on a transverse wave is 12.0cm. If cycles of the wave pass a fixed point in one sec

11Alexandr11 [23.1K]

Answer:

0.12m/s

Explanation:

v=λf

Given that, λ = 12cm = 0.12m

T = 1second

(A period T is the time required for one complete cycle of vibration to pass a given point)

frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz

therefore, v= 0.12 × 1 = 0.12m/s

8 0

2 years ago

Suppose you drive an average of 25 miles a day, for a total of 60 years in a lifetime. The total distance you would have covered

Ahat [919]

The total distance you would have covered is equivalent to going around the earth 22 times.

<h3>Total distance traveled</h3>

The total distance traveled at a given speed and time of motion is calculated as follows;

Distance = speed x time

$Distance = \frac{25 \ miles}{day} \times \frac{365 \ days}{1 \ year} \times 60 \ years\\\\Distance = 547, 500 \ miles$

<h3>Distance round the Earth</h3>

The distance round the earth or circumference of the earth of the earth has been given as 25,000 miles

<h3>Number of times round the earth</h3>

$n = \frac{547,500}{25,000} \\\\ n = 21.9 \ times \ \approx 22 \ times$

Thus, the total distance you would have covered is equivalent to going around the earth 22 times.

Learn more about distance here: brainly.com/question/17273444

3 0

2 years ago

An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-

Gnoma [55]

Answer:

$1.357\times 10^{-12}$

Explanation:

Relevant Data provided

Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2

Distance which indicates d = 1.50 x 10^-3 m

Voltage which indicates V = 12 V

According to the requirement, the computation of value of its capacitance is shown below:-

$Capacitance, C = \frac{\epsilon oA}{D}$

$= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}$

= $1.357\times 10^{-12}$

Therefore for computing the capacitance we simply applied the above formula.

4 0

3 years ago

Why is speed greater at the end of a ramp than it is at the beginning of a ramp?

UkoKoshka [18]

At the begging nothing is happening, but when you push it, it gains speed that is why the end of a ramp is faster.

5 0

3 years ago