What force on a 936kg elevator that is falling freely at 9.8 m/s^2

To calculate acceleration you must know both the objects velocity and_____

amm1812

You need to know the time as well.

8 0

3 years ago

What is Agent Object Notation? ( For Force Diagrams. )

kobusy [5.1K]

F sub n is an example.

6 0

3 years ago

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe

kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0

3 years ago

Answer the following question

Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0

3 years ago

In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b

V125BC [204]

The total average velocity $v=+1.41 m/s$ (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
$v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}$
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion ( $S_1=6.30 km=6300 m$ , due west) and of the displacement in the second part of the motion ( $S_2$ , due east).
-The total time taken t is the time taken for the first part of the motion, $t_1$ , and the time taken for the second part of the motion, $t_2$ . $t_1$ can be found by using the average velocity and the displacement of the first part:
$t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s$

$t_2$ , instead, can be written as $\frac{S_2}{v_2}$ , where $v_2=-0.630 m/s$ is the average velocity of the second part of the motion (with a negative sign, since it is due east).

Therefore, we can rewrite the initial equation as:
$v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }$
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
$S_2=-844 m=-0.844 km$

4 0

3 years ago

What force on a 936kg elevator that is falling freely at 9.8 m/s^2​

What force on a 936kg elevator that is falling freely at 9.8 m/s^2