Answer:
i took g = 9.8m/s
A. 1.16secs
B. 2.32secs
C. 6.57m
D. 57.91m
Explanation:
A. How long does the Missile take to reach ot peak?
Time taken (t) =( U²Sin (angle) )/g
u = initial velocity = 25m/s
angle given = 30°
g = acceleration due to gravity = 9.8m/s²
t = U² x Sin (angle) / g
t = 25² x Sin(30)/9.8
t = 1.61secs
B. How long is the missile in the air in total?
T = 2t
T = 2 x 1.61 = 2.32 secs
C. what maximum Height does the missile reach?
- Maximum height = U²Sin²(angle) / 2g
- M.H =25² x Sin(30)² / 2 x 9.8
- M.H=6.57m
- Maximum height= 6.57m
D. How far does the missile travel Horizontally?
- Range = U²2Sin(angle)/g
- Range = 25² x 2 x Sin(30) / 9.8
- Range = 57.91m
Answer:Dr. Hess' most significant contribution to the plate tectonic theory began in 1945 when he was the commander of the U.S.S. ... In the paper Hess described how hot magma would rise from under the crust at the Great Global Rift. When the magma cooled, it would expand and push the tectonic plates apart.
Explanation:
Answer:
Explanation:
Let the luminosity of the star be I and luminosity of the sun be Isun.
2.4 billion light years = 2.4 x 10⁹ light years .
brightness = luminosity / (distance)²
Given Sun would have to be viewed from a distance of 1300 light-years to have the same apparent magnitude as 3C 273 so
For the sun
brightness = Isun / (1300 light years )²
For star
brightness = I / (2.4 x 10⁹ light years )²
Both these brightness are same
Isun / (1300 light years )² = I / (2.4 x 10⁹ light years )²
I = Isun x (2.4 x 10⁹ light years )² / (1300 light years )²
= Isun x 3.4 x 10¹² .
Answer:
The total frictional force is 358.0 newtons
Explanation:
Power is the amount of average work (W) an object does on a period of time (Δt):
Remember average work is average force (F) times displacement (Δs):
but displacement over time is average speed , then:
(1)
That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:
Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308 = , then:
Answer:
19.12 m/s.
Explanation:
The following data were obtained from the question:
Final velocity (v) = 0 m/s
Breaking Acceleration (a) = – 4.57 m/s²
Distance (s) = 40 m
Initial velocity (u) =.?
The initial velocity of the car can be obtained as illustrated below:
v² = u² + 2as
0² = u² + (2 × –4.57 × 40)
0 = u² + (– 365.6)
0 = u² – 365.6
Collect like terms
u² = 0 + 365.6
u² = 365.6
Take the square root of both side
u = √365.6
u = 19.12 m/s
Therefore, the car was initially moving at 19.12 m/s.