Ask question

Ask question

All categories

3 years ago

6

olivia is taking pictures of the ocean on the edge of a cliff when she accidentally drops her phone. The phone hits the water at

a vertical velocity of 15.75m/s and landed 16m away from the base of the cliff. how tall is the cliff?

1 answer:

In-s [12.5K]3 years ago

3 0

The cliff is 12.71m tall.

Why?

Before solving the problem, we must remember that the phone was dropped, it means that there is no initial velocity, also, the acceleration acting on the phone is the acceleration due to gravity.

So, to find how tall is the cliff, we can use the following equations:

$y=y_o+v_o*t-\frac{1}{2}g*t^{2}\\\\v_f=v_o+g*t$

$t=\frac{v_f}{g} \\\\t=\frac{-15.75\frac{m}{s}}{-9.81\frac{m}{s^{2}}}=1.61s$

Now that we know the time, we can calculate how tall is the cliff:

$y=y_o+v_o*t+\frac{1}{2}g*t^{2}\\\\y=0+0*1.61s-\frac{1}{2}*(-9.81\frac{m}{s^{2}})*(1.61s)^{2}\\\\y=12.71m$

Hence, we have that the cliff is 12.71m tall.

Have a nice day!

You might be interested in

How does the thermoflask prevent heat loss

riadik2000 [5.3K]

Answer:

The way that the flask is built it has 3 protective layers.... the inside layer to keep the heat in, the outside layer to reflective the cold, and a vacuum layer, which is an empty layer that limits conduction and convection

Explanation:

3 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

andreyandreev [35.5K]

Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required.

25 kg * 8 m = work = 100 kg * 2 m

7 0

2 years ago

A flare is shot from a snowmobile with an initial velocity of 4.3 m/s at 90*, while the snowmobile is moving at 8.5 m/s at 0.0*.

Sedaia [141]

The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.

He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.

When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.

The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about 9.53 m/s, at some angle between straight up
and straight north.

The angle above horizontal is the angle that has a tangent of 4.3/8.5 .
I'll let you work out the angle.

8 0

2 years ago

Waves combine to produce a smaller or zero-amplitude wave in a process called

vampirchik [111]

ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!

6 0

3 years ago

Read 2 more answers