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3 years ago

A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car

Physics

1 answer:

lisabon 2012 [21]3 years ago

5 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

Hope this helps you

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Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste

Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

$W=0$

We need to calculate the system's potential energy

Using thermodynamics first equation

$\Delta U=W-\Delta E$

Put the value into the formula

$\Delta U=0-147$

$\Delta U=-147\ J$

Hence, The system's potential energy is -147 J.

5 0

3 years ago

List at least three things that can happen as plate spread apart.

Nookie1986 [14]

Centre of Mass then axis of rotation and then moment of inertia. This was the toughest question for your level... happy to help ^_^. It was purely experimental question.

8 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0

3 years ago

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

3 years ago

in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

BabaBlast [244]

Answer:

$\boxed{3 \ Significant \ figures}$

Explanation:

Total spectators = 5000

Counted by the groups of ten, So at last the result will be:

=> 5000/10 = 500

Significant figures in 500 are 3

8 0

3 years ago