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3 years ago

A stone is dropped from a bridge abd it hits the water 2.2 seconds later how high is the bridge above the water

Physics

1 answer:

Bess [88]3 years ago

8 0

Answer:

h = 23.716 m

Explanation:

Given that,

The time taken by the stone to hit the water is, t = 2.2 s

Height of the bridge above the ground, h = ?

The distance that the body will fall through the time is given by the formula

S = 1/2 gt² m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

S = 1/2 x 9.8 m/s² x (2.2 s)²

= 23.716 m

Therefore, the height of the bridge from the surface of the water is h = 23.716 m

You might be interested in

a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec

Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound $s$ is given by a relation between the distance $d$ and the time $t$ :

$s=\frac{d}{t}$ (1)

Where:

$s=330 m/s$ is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

$d=412.5 m-x$ since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance $x$ towards the cliff

$t=\frac{2.2 s}{2}=1.1 s$ Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate $d$ and then find $x$ :

$d=st$ (2)

$412.5 m-x=(330 m/s)(1.1 s)$ (3)

Finding $x$ :

$x=49.5 m$ This is the distance at which the hunter is from the cliff.

3 0

3 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago

A 77.0 kg firefighter climbs a flight of stairs 8.0 m high. how much work is required? 0 incorrect: your answer is incorrect. j

Alex73 [517]

<span>6160 joules

to lift 1 newton 1 metre requires 1 joule
there are 10 newtons in one kilo

so 77(kg) x 8 (metres) x 10 (newtons/kilo) = 6160 joules</span>

5 0

4 years ago

The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg

shepuryov [24]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷

Explanation:

Given:

Mass of α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷

4 0

3 years ago

A topic list is a list of important content and material which will be covered on a test. Please select the best answer from t

Iteru [2.4K]

Sounds T to me. My teacher used to give those out.

6 0

3 years ago

Read 2 more answers