Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>

the velocity is Zero when the projectile reach in the maximum altitude:

When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>

R=Range


**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile

2B=60°
B=30°
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.
First we calculate the fuel consumed by each car in a year
Fuel consumed=6990/21.4=326.63 gallon
Now we calculate the amount of fuel consumed by 40 million cars in a year
Fuel consumed=326.63*40*10^6=13065.42 million gallon,
Now we can calculate the annual production of carbon dioxide in the USA
CO2 production rate=9.50*13065.42=124121.49*10^6 kg
Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg
Answer:
The distance travel by block before coming to rest is 0.122 m
Explanation:
Given:
Mass of block
kg
Initial speed of block

Final speed of block

Coefficient of kinetic friction 
Ramp inclined at angle
28.4°
Using conservation of energy,
Work done by frictional force is equal to change in energy,

Where 



m
Therefore, the distance travel by block before coming to rest is 0.122 m