PHYSICS HIGH SCHOOL

A soccer ball is released from rest at the top of a grassy incline. After 4.1 seconds, the ball travels 43 meters and 1.0 s after this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant? Express your answer using two significant figures.

Answers

Answer 1

Answer:

The acceleration of the ball as it moves down the grassy incline, if constant, is 5.1 m/s²

Explanation:

Initial velocity, u = 0m/s

Acceleration = ?

Vertical distance covered as at t=4.1s, H = 43m

Using the equations of motion,

H = ut + 0.5at²

43 = 0 + 0.5a (4.1)²

a = 43/(0.5×4.1²) = 5.1 m/s²

Hope this Helps!!

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Answer:

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Explanation:

A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?

Answers

75 km/hr, so in other words 75 km per hour.

7200 s / 60 = 120 min
120 min / 60 =2 hr
so 150 km / 2 hr = 75 km per hour

Answer:

The speed of the car is $\boxed{75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}}$ or $\boxed{20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The distance covered by the car is $150\,{\text{km}}$ .

The time taken by the car to cover the distance is $7200\,{\text{s}}$ .

Concept:

The speed of a car is the ratio of the distance covered by the car to the time taken by the car in covering the distance.

The expression for the speed of the car is:

$v = (d)/(t)$ …… (1)

Here, $v$ is the speed of the car, $d$ is the distance covered and $t$ is the time taken by the car.

Substitute the value of distance and time in the above expression.

$\begin{aligned}v &= \frac{{150\,{\text{km}}}}{{7200\,{\text{s}}\left( {\frac{{1\,{\text{hr}}}}{{60 * 60\,{\text{s}}}}} \right)}}\n&= \frac{{150}}{2}\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\n&= 75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}} {{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\n\end{aligned}$

The speed of the car in terms of ${{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ can be expressed as.

$\begin{aligned}v&=75\,{{{\text{km}}} \mathord{\left/{\vphantom {{{\text{km}}}{{\text{hr}}}}} \right.\kern-\nulldelimiterspace} {{\text{hr}}}}\left( {\frac{{1000\,{\text{m}}}}{{3600\,{\text{s}}}}} \right)\n&=75 *\frac{5}{{18}}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\n&= 20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\n\end{aligned}$

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Answer Details:

Grade:Middle School

Chapter:Speed and distance

Subject:Physics

Keywords:Speed, velocity, distance, time, distance of 150 km, in 7200 s, speed of the car, ratio of distance, rate of change of distance.

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