Ask question

Ask question

All categories

3 years ago

5

FOR 50 POINTS! -- Your car is parked at the high school. After school you take off and travel for 27 km towards Iowa City. At th

is point, you realize you forget your wallet and travel back to Tipton for 15 km before you find your wallet was just on the floor. You turn back and head towards Iowa city for 36 km.
What is your displacement?

2 answers:

Dovator [93]3 years ago

6 0

Answer:

36km

Explanation:

Im pretty sure displacment is the start and finish in a straight line

Vilka [71]3 years ago

4 0

Answer:

27-15= 12

Explanation:

The official displacement formula is as follows:

s = sf – si

s = displacement

si = initial position

sf = final position

I really hope this is right I felt really bad for getting it wrong last time and wanted to make it up to ya lol

You might be interested in

B. On a separate sheet of paper, describe the different ways of generating electric power.

Afina-wow [57]

Answer:

These all different sources of energy add to the store of electrical power that is then sent out to different locations via high powered lines. It is the energy from the sun that is harnessed using a range of technologies such as solar heating, solar architecture, photovoltaics, and artificial photosynthesis.

Hope it helps PLS MARK ME AS BRAINLIST I BEG YOU thanks :)

4 0

2 years ago

leva [86]

Answer:

Option C

Maximum potential energy is at point R.

Explanation:

Potential energy is a product of mass, acceleration due to gravity and height ie

PE=mgh where PE is the potential energy, m is mass of an object, g is acceleration due to gravity whose value is normally taken as 9.81 and h is height. Since at point R we have the maximum height, the potential energy will be highest at this point.

3 0

3 years ago

A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

emmainna [20.7K]

Complete Question

A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The pumpkin's vertical displacement is $H = 27 .67 \ m$

The pumpkin's vertical velocity when it hits the ground is $v_v__{f}} = 23.298 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $v_h = 4 m/s$

The horizontal distance traveled is $d = 9.5 \ m$

The horizontal distance traveled is mathematically represented as

$S = v_h * t$

Where t is the time taken

substituting values

$9.5 = 4 * t$

=> $t = \frac{9.5}{4}$

$t = 2.38 \ sec$

Now the vertical displacement is mathematically represented as

$H = v_v t + \frac{1}{2} a_v t^2$

now the vertical velocity before the throw is zero

So

$H = 0 + \frac{1}{2} (9.8) * (2.375)^2$

$H = 27 .67 \ m$

Now the final vertical velocity is mathematically represented as

$v_v__{f}} = v_v + at$

substituting values

$v_v__{f}} = 0 + (9.8)* (2.375)$

$v_v__{f}} = 23.298 \ m/s$

7 0

3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago

What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr

yKpoI14uk [10]

Check the attached file for the solution.

8 0

3 years ago