What science projects should I do? Please just choose something basic!

What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

3 years ago

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0

3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

3 years ago

In the year 2081 in a shipping port on the moon, workers for Ore-Space, Inc., hoist a 500.0 kg hunk of anorthosite moon rock by

Naddik [55]

Answer:

2,800 n

Explanation:

hope this helps, have a nice day/night! :D

7 0

3 years ago

A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning

Lesechka [4]

Answer:

speed of current is 5 mile/hr

Explanation:

GIVEN DATA:

speed of motorboat = 15 miles/hr relative with water

let c is speed of current

15-c is speed of boat at upstream

15+c is speed of boat at downstream

we know that

travel time=distance/speed

$\frac{10}{15-c} +\frac{10}{15+c} = 1.5$

150+10c+150-10c=1.5(15-c)(15+c)

300=1.5(225-c^2)

300=337.5-1.5c^2

200=225-c^2

c^2=25

c = 5

so speed of current is 5 mile/hr

6 0

3 years ago