Pressure increases with depth towards the center of earth. In which layer would you expect pressure to be the greatest.

At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

3 years ago

TECHNOLOGY A Lamborghini Aventador accelerates from rest to 100 km / h in 2.9 s. The mass of the car is 1575 kg Calculate the po

DerKrebs [107]

At 100 km/hr, the car's kinetic energy is

KE = (1/2) (mass) (speed)²

KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²

KE = (787.5 kg) (27.78 m/s)²

KE = 607,639 Joules

In order to deliver this energy in 2.9 seconds, the engine must supply

(607,639 J / 2.9 sec) = 209,531 watts

<em>Power = 281 HP</em>

6 0

3 years ago

vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di

pantera1 [17]

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

6 0

3 years ago

Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its

antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$ given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem $\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

$\mathrm{x}=\frac{\sqrt{2}}{4} L$$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$I_x=I_8+M x^2$

$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$ , Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

5 0

1 year ago

Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550

Arisa [49]

Answer:

q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2 (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

w_poly = R*(T_f - T_i)/(1-n)

w_poly = 0.189*(350 - 400)/(1-1.2)

w_poly = 47.25 KJ/kg

-Hence,

q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

q_poly = 14.55 KJ/kg

4 0

3 years ago