The diagram shows a skydiver at different points of her jump. At what point would her

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

8 0

3 years ago

The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids

BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

$100\times5\times60= m(1700)20+(0.63-m)(4200)20$

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0

3 years ago

If the moon were twice as far from years as it is now the following would be true

zubka84 [21]

The question is incomplete.

The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.

Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.

Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.

7 0

3 years ago

Why is vehicle systems forensics useful today? Cars are less computerized than before More people own cars Devices like smartpho

kvasek [131]

Answer:

Devices like smartphones can interface with cars and leave evidence

Explanation:

Vehicle system forensic relates to digital data stored in a vehicles system.

Bluetooth connection times can be used to figure out at what time the owner was near his car. e.g. between a smart-watch and car infotainment system

6 0

3 years ago

I WILL GIVE BRAINLIST just please help with these problems 3

statuscvo [17]

Answer:

16. 68.18 Km/h

17. 3 miles.

Explanation:

16. Determination of the speed

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed is simply defined as the distance travelled with time. Mathematically, it is expressed as:

Speed = Distance / time

With the above formula, we can obtain the speed as follow:

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed = Distance /time

Speed = 150 / 2.2

Speed = 68.18 Km/h

17. Determination of the distance.

Speed = 3 mph

Time = 1 hour

Distance =?

Speed = Distance /time

3 = distance / 1

Distance = 3 miles

3 0

2 years ago