A cannon ball was launched at a degree of 20° with an initial velocity of 20m/s.1. What is the distance traveled by the object after 6s if it reach a final velocity of 30m/s?2. Calculate the range it travels.
1. A cannon ball was launched at a degree of 20° with an initial velocity of 20m/s.1. What is the distance traveled by the object after 6s if it reach a final velocity of 30m/s?2. Calculate the range it travels.
Answer:
1.630kh
2.2568km
Explanation:
Hope it is help paki vote
2. what is the kinetic energy of an object that has a mass of 100kg and a velocity' of 20m/s²
Answer:
KE=kinetic energy
m=mass of a body
v=velocity of a body
Explanation:
KE=2000 j
3. what is the horizontal range of the object if the velocity is 20m/s and the angle is 25 degrees.
Answer:
31.26 m
Step-by-step explanation:
Formula for horizontal range is
R = u² sin 2θ / g
u = initial velocity
θ = angle of trajectory
g = acceleration by gravity
R = u² sin 2θ / g
R = (20)² sin 2(25) / 9.8
R = 400sin(50) / 9.8
R ≈ 31.26 m
4. an object with an initial velocity of 20m/s accelerates uniformly at 5m/s² in the direction of its motion for distance of 10m. What is the final velocity of the object? Show your solution.
Explanation:
An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s2 for 40.0 s. What is its final velocity?
Strategy
First, we identify the knowns:
\[{v}_{0}=70\,\text{m/s,}\,a=-1.50{\,\text{m/s}}^{2},t=40\,\text{s}\]
.
Second, we identify the unknown; in this case, it is final velocity
\[{v}_{\text{f}}\]
.
Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using (Figure),
\[v={v}_{0}+at\]
.
Solution
[reveal-answer q=”287818″]Show Answer[/reveal-answer]
[hidden-answer a=”287818″]Substitute the known values and solve:
\[v={v}_{0}+at=70.0\,\text{m/s}+(-1.50\,\text{m/}{\text{s}}^{2})(40.0 s)=10.0 m/s.\]
(Figure) is a sketch that shows the acceleration and velocity vectors.[/hidden-answer]
Figure shows airplane at two different time periods. At t equal zero seconds it has velocity of 70 meters per second and acceleration of -1.5 meters per second squared. At t equal 40 seconds it has velocity of 10 meters per second and acceleration of -1.5 meters per second squared.
Figure 3.19 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note the acceleration is negative because its direction is opposite to its velocity, which is positive.
Significance
The final velocity is much less than the initial velocity, as desired when slowing down, but is still positive (see figure). With jet engines, reverse thrust can be maintained long enough to stop the plane and start moving it backward, which is indicated by a negative final velocity, but is not the case here.
5. • An object with an initial velocity of 25m/s accelerates uniformly at 5m/s² in the direction of its motion for a distance of 20m. What is the final velocity of the object? need helpno joke pls tysm
Answer:
28.72 m/sStep-by-step explanation:
Formula:
[tex] \\ \tt{}V_{f} {}^{2} = V_{i} {}^{2} + 2ad\\ \\[/tex]
[tex] \tt{}V_{f} [/tex]= final velocity = ?
[tex] \tt{}V_{i} [/tex] = initial velocity = 25m/s
a = acceleration = 5m/s²
d = displacement = 20m
[tex]\tt{}V_{f} {}^{2} =25 {}^{2} + 2(5)(20)[/tex]
[tex]\tt{}V_{f} {}^{2} =625 + 2(100)[/tex]
[tex]\tt{}V_{f} {}^{2} =625 + 200[/tex]
[tex]\tt{}V_{f} {}^{2} =825[/tex]
[tex]\tt{}V_{f} = \sqrt{825} [/tex]
[tex]\tt{}V_{f} =28.7228132327[/tex]
[tex] \boxed{ \large \boxed{\tt{}V_{f} =28.72}}[/tex]
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#BrainliestBunch
Stay safe and God bless you!
[tex] \gamma \\ \huge \boxed{ \ddot \smile}[/tex]
6. an object is located 60m above the ground and was launched horizontally with velocity of 28m/s
Answer:
The numerical information in both the diagram and the table above illustrate identical points - a projectile has a vertical acceleration of 9.8 m/s/s, downward and no horizontal acceleration. This is to say that the vertical velocity changes by 9.8 m/s each second and the horizontal velocity never changes
7. with an initial velocity of 20m/s,a car accelerated at 8m/s² for 12 seconds
Answer:
Question: What is acceleration?
Answer: Change in velocity per second.
Change in velocity: 20 m/s - 8 m/s = 12 m/s
How long did it take: 5 s
Acceleration: 12 m/s / 5 s = (you take it from here)
8. A 20-kg object travelling 20ms collides heladion with an 18-kg object travelling at 17 m/s. If they were locked together after the collision, what is their velocity?
Answer:
20 20 18 17
Explanation:
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9. An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. 1. What is the horizontal range? 2. What is the magnitude of the velocity of the object just before it hits the ground?
Calculating the horizontal range on projectile motion.
Given
Velocity (Vₒ) = 20 m/sAngle of Trajectory (θ) = 25°Horizontal Range (x) = ?In finding the formula for the horizontal range, we need to find the variable of time travelled of the projectile.
We need to derive first the formula for vertical range since the direction of the projectile is upwards.
The formula with time in the vertical range is.
y = yₒ + Vₒy t + 1/2gt²
- where g is a negative value of the acceleration of gravity ( -9.8 m/s² )
Derive time to the formula of the vertical range. Plug in the given values.
Remember that
Vₒy = Vₒ sin θ
0 = 0 + ( Vₒ sin θ ) t + 1/2gt²
0 = t ( Vₒ sin θ + 1/2gt)
Vₒ sin θ + 1/2gt = 0
1/2gt = -Vₒ sin θ
t = 2 Vₒ sin θ / |g|
Now that we already found the formula for time, next is we have to substitute the formula of time for the formula of horizontal range.
The formula is
x = (Vₒ cos θ)(t)
x = Vₒ cos θ 2 Vₒ sin θ / |g|
We can also conclude that
2 sin θ cos θ = sin 2θ
Then replace it to the formula
x = Vₒ² sin 2θ / |g|
Then we can plug in the values to the formula
x = Vₒ² sin 2θ / |g|
x = (20)²(sin 2(20)) / |-9.8 m/s|
x = (400)(sin 40) / 9.8 m/s
x = 257.1150 / 9.8 m/s
x = 26.24 m
#CarryOnLearning
10. An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal. (a) What is the maximum height reacted by the object? (b) What is the total time (between launch and touching the ground) of the object? And (c) What is the horizontal range of the object.
An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal.
Given:
v¡ = 45 m/sθ = 25°g = 9.8m/s²a) The maximum height reached by the object.[tex]d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]
[tex]d_{y} = \displaystyle\frac{(45 \: m/s \times \sin 25\degree)^{2}}{2(9.8 m/s²)}[/tex]
[tex]d_{y} = 18.45 \: m[/tex]
Answer: [tex]\: \: \sf\large\blue{18.45 m}[/tex]b) The time of the object.[tex]t = \displaystyle\frac{2v_{i} \sin\theta}{g}[/tex]
[tex]t = \displaystyle\frac{2(45 \: m/s) \sin 25\degree}{9.8 m/s²}[/tex]
[tex]t = 3.88 \: s[/tex]
Answer: [tex]\: \: \sf\large\blue{3.88 s}[/tex]c) The horizontal range of the object.[tex]R = \displaystyle\frac{v_{i} \: ^{2} \sin2\theta}{g}[/tex]
[tex]R = \displaystyle\frac{(45 \: m/s)^{2} \sin 50\degree}{9.8 m/s²}[/tex]
[tex]R = 158.29 \: m[/tex]
Answer: [tex]\: \: \sf\large\blue{158.29 m}[/tex]#BrainliestBunch
11. an object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal
The correct answer are 3.64 m
12. An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal. (a) What is the maximum height reacted by the object? (b) What is the total time (between launch and touching the ground) of the object? And (c) What is the horizontal range of the object.
An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal.
Given:
v¡ = 45 m/sθ = 25°g = 9.8m/s²a) The maximum height reached by the object.[tex]d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]
[tex]d_{y} = \displaystyle\frac{(45 \: m/s \times \sin 25\degree)^{2}}{2(9.8 m/s²)}[/tex]
[tex]d_{y} = 18.45 \: m[/tex]
Answer: [tex]\: \: \sf\large\blue{18.45 m}[/tex]b) The total time of the object.[tex]t = \displaystyle\frac{2v_{i} \sin\theta}{g}[/tex]
[tex]t = \displaystyle\frac{2(45 \: m/s) \sin 25\degree}{9.8 m/s²}[/tex]
[tex]t = 3.88 \: s[/tex]
Answer: [tex]\: \: \sf\large\blue{3.88 s}[/tex]c) The horizontal range of the object.[tex]R = \displaystyle\frac{v_{i} \: ^{2} \sin2\theta}{g}[/tex]
[tex]R = \displaystyle\frac{(45 \: m/s)^{2} \sin 50\degree}{9.8 m/s²}[/tex]
[tex]R = 158.29 \: m[/tex]
Answer: [tex]\: \: \sf\large\blue{158.29 m}[/tex]#BrainliestBunch
13. an object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a)What is the maximum height reached by the object?b) What is the total flight time ( between launch and touching the ground) of the object?
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
Given:
v¡ = 20 m/sθ = 25°g = 9.8m/s²a) The maximum height reached by the object.[tex]d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]
[tex]d_{y} = \displaystyle\frac{(20 \: m/s \times \sin 25\degree)^{2}}{2(9.8 m/s²)}[/tex]
[tex]d_{y} = 3.65 \: m[/tex]
Answer: [tex]\: \: \sf\large\blue{3.65 m}[/tex]b) The total flight time of the object.[tex]t = \displaystyle\frac{2v_{i} \sin\theta}{g}[/tex]
[tex]t = \displaystyle\frac{2(20 \: m/s) \sin 25\degree}{9.8 m/s²}[/tex]
[tex]t = 1.72 \: s[/tex]
Answer: [tex]\: \: \sf\large\blue{1.72 s}[/tex]#BrainliestBunch
14. An object is launched at a velocity of 30 m/s in a direction making an angle of 35° upward with the horizontal. What is the total flight time (between launch and touching the ground) of the object? A. 2s B. 1.9s C. 3.51s D. 3.2s
Answer:
B.
Solution:
Vy = 30sin35
Vy = 17.207m/s
t = 2(17.207/9.8)
t = 3.5116s
15. An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. 1. What is the magnitude of the velocity of the object just before it hits the ground?
To find the magnitude of the velocity of the object just before it hits the ground, we can use the conservation of mechanical energy principle. The initial kinetic energy plus the initial potential energy equals the final kinetic energy plus the final potential energy. Since the object is at the same height when it hits the ground, the initial and final potential energies are equal, and we can focus on the kinetic energies.
First, let's find the initial horizontal (Vx) and vertical (Vy) components of the velocity:
Vx = V * cos(θ)
Vy = V * sin(θ)
where V = 20 m/s and θ = 25°.
Vx = 20 * cos(25°)
Vy = 20 * sin(25°)
Now, we can find the final vertical component of the velocity (Vy_final) using the conservation of mechanical energy principle:
Vy_final^2 = Vy^2 + 2 * g * h
Since h = 0 (the object is at the same height when it hits the ground), we have:
Vy_final^2 = Vy^2
Taking the square root of both sides:
Vy_final = Vy
Now, we can find the magnitude of the final velocity (V_final) using the Pythagorean theorem:
V_final = sqrt(Vx^2 + Vy_final^2)
Let's calculate the values:
Vx = 20 * cos(25°) ≈ 18.08 m/s
Vy = 20 * sin(25°) ≈ 8.47 m/s
Vy_final = Vy ≈ 8.47 m/s
V_final = sqrt(18.08^2 + 8.47^2) ≈ 20 m/s
The magnitude of the velocity of the object just before it hits the ground is approximately 20 m/s.
16. 2. Object A with a mass of 3kg and a velocity of 20 m/s collide head-on with object B having a mass of 3 kg and a velocity of -30 m/s. If the collision is perfectly elastic and the first object bounces back with a velocity of -30 m/s, what will be the velocity of the second object after collision? GIVEN: m1=3kg v1=20m/s v2=-30 m/s UNKNOWN=v2' m2=3kg v1' = -30m/s FORMULA=m1v1+m2v2=m1v1'+m2v2
BAKA NAMAN OHH PA BRAINLIST AT LIKE STAR NAMAN SANA MAKATULONG
17. An object that is launched from the top of a tower is dense enough that air resistance can be ignored. The object's initial velocity has components (vx=60m/s, vy=20 m/s) What are the components of the object's velocity 5 seconds after launch?
Answer:
Objects in Flight
Objects in Flight Due to the shot's high mass-density, however, forces such as air friction and the downward pull of gravity affect its flight less than a softball's. In other words, the shot will travel farther than the softball, because its mass makes it less vulnerable to air friction and gravity.
18. how long does it take to increase car's velocity fron 20m/s to 40m/s if a car can accelerate at a rate of 4m/s²? 2.compute the initial velocity of an object that is accelerating at 10m/s² with the velocity of 200 m/s in 10 seconds? 3.how far does moving object travel that accelerates at a rate of 5m/s² with a change of velocity from 5m/s to 10m/
Answer:120 m.Step-by-step explanation:
5m/s² means that the velocity increase each second in 5m/s. So 4 s of that acceleration would increase the speed (in m/s) from 20 to 40. (Speed increase each second in 5 m/s. We need an increase of 20 m/s.)
Since the acceleration is uniform during does 4 s, we can use the simple average speed of 30 m/s. 30 m/s * 4 s = 120 m.
19. A 25 kg object falling towards earth has a velocity of 8.5 m/s when it is 100m above the ground. What will be its velocity when it is 20m above the ground?
Answer:
25 kg object is falling towards the earth with a velocity of 8.5 m/s when it is 100 m above the ground. What will be its velocity when it is 20 m above the ground it energy is conserved? ½mv² + mgh = ½mv² + mgh. ½(25.0kg)(8.5 m/s)² + (25.0 kg)(9.8 m/s²)(100m) = ½(25.0kg)(v)² + (25.0 kg)(9.8 m/s²)(20m)
20. An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range of the object?
Answer:
Given : An object is launched at a velocity of 36m/s in a direction making an angle of 29⁰ upward with the horizontal
To Find :
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range
c) What is the magnitude of the velocity of the object just before it hits the ground?
Solution:
Horizontal Velocity = 20cos25° = 18.13 m/s
Vertical Velocity = 20Sin25° = 8.45 m/s
Horizontal Velocity remains constant
Vertical velocity becomes 0 at max height
V = u + at
a = - g = -9.8 m/s²
=> 0 =8.45 - 9.8t
=> t =0.86 sec
Total Time of flight = 2 * 0.86 = 1.72 sec
V² - U² = 2as
Max height = (v² - u²)/2a = -(8.45)²/(2*(-9.8))
= 3.64 m
Horizontal range = 1.72 x 18.13 = 31.18 m
magnitude of the velocity of the object just before it hits the ground
= 20 m/s ( based on conservation of energy )
Initial KE + initial PE = Final KE + Final PE
Initial and final PE = 0
Initial KE = Final KE
Hence Magnitude of velocity will be same
21. An object qith an initial velocity of 20m/s accelerates uniformly at 3m/s² in the direction of its motion for a distance of 10m. what is the finak velocity of the object? Given: RTF: Formula: Solution:pa sagot mo ayos ty^^
Answer:
yn na po pa brainliest answer nmn po
Explanation:
Step-by-step:
22. A car with an initial velocity of 20m/s accelerate at rate of 80m/s for 5 seconds what will be the final velocity
final velocity would be this... let it be shown in the acceleration formula, a= vf-vi/t= Δv/sec.
a= (80m/s-20m/s)/5s= 60m/s/5s. The final velocity would be 80m/s.
23. An object was launched from horizontal with a velocity of 10.5 m/s. If it takes 1.25 seconds for the object to land on the ground. Calculate the following: launching height and Range
Answer:
launch height:
h = Vo*t + (1/2)*g*t^2
h = (10.5m/s)*(1.25s) + (1/2)*(9.8m/s2)*(1.25s)2
h = 16.675 m
Range:
R = Vo*t + (1/2)*a*t^2
R = (10.5m/s)*(1.25s) + (1/2)*(0m/s2)*(1.25s)2
R = 13.125 m
24. u & A 2010 object travelling at 20m/s collides head B-kg object travelling at 17016 If they were together after on with an locked the collisions what is their velocity ?solution?
Answer:
17,036
solution: 20 + 17016=17,036
Step-by-step explanation:
sana po makatulong.
25. An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal. (a) What is the maximum height reacted by the object? (b) What is the total time (between launch and touching the ground) of the object? And (c) What is the horizontal range of the object.
An object is launched at a velocity of 45 m/s in a direction making an angle of 25° upward with the horizotal.
Given:
v¡ = 45 m/sθ = 25°g = 9.8m/s²a) The maximum height reached by the object.[tex]d_{y} = \displaystyle\frac{(v_{i} \sin \theta)^{2}}{2g}[/tex]
[tex]d_{y} = \displaystyle\frac{(45 \: m/s \times \sin 25\degree)^{2}}{2(9.8 m/s²)}[/tex]
[tex]d_{y} = 18.45 \: m[/tex]
Answer: [tex]\: \: \sf\large\red{18.45 m}[/tex]b) The total time of the object.[tex]t = \displaystyle\frac{2v_{i} \sin\theta}{g}[/tex]
[tex]t = \displaystyle\frac{2(45 \: m/s) \sin 25\degree}{9.8 m/s²}[/tex]
[tex]t = 3.88 \: s[/tex]
Answer: [tex]\: \: \sf\large\red{3.88 s}[/tex]c) The horizontal range of the object.[tex]R = \displaystyle\frac{v_{i} \: ^{2} \sin2\theta}{g}[/tex]
[tex]R = \displaystyle\frac{(45 \: m/s)^{2} \sin 50\degree}{9.8 m/s²}[/tex]
[tex]R = 158.29 \: m[/tex]
Answer: [tex]\: \: \sf\large\red{158.29 m}[/tex]#BrainliestBunch
26. If a car, with an initial velocity of 20m/s, accelerates at a rate of 5m/s² for 3 seconds, what will its final velocity be
Formula:
Vf = Vi + (a x t)
Solution:
Vf = Vi + (a x t)
Vf = 20 m/s + ((5 m/s^2)(3 s))
Vf = 20 m/s + 15 m/s
Vf = 35 m/s
The final velocity is 35 m/s.
27. how much is the kinetic energy of an object moving downward the building with a velocity of 20m/s and has a mass of 200kg?
KINETIC ENERGY
» How much is the kinetic energy of an object moving downward the building with a velocity of 20 m/s and has a mass of 200 kg?
Given:m (mass) = 200 kgv (velocity) = 20 m/sSolution:KE = ½ × m × vKE = ½ × 200 × 20²KE = ½ × 200 × 400 KE = 40,000 Answer:40,000 J (joules)_______________◇_______________
28. 60° with an initial velocity of 20m/s
Answer:
12jejeeeeejejejejejjrdjdjdjdjjddjjdddj
29. An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object? d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
C.
Explanation:
carry on learning
just ignore I'f I'm wrong po:)
pa Brainlies I'f I'm correct thank you po:)
30. A football was launched in the air. The angle of the football from the player was 75° with a velocity of 20m/s. What are the vertical andhorizontal components of the football? A. Horizontal: 5.18m/s ! Vertical: 19.32 m/sB. Horizontal: 70.48m/s 1 Vertical 19.32m'sC Horizontal: 70.48m/s | Vertical: 25.65m/s
Answer:
A. Horizontal: 5.18 m/s, Vertical: 19.32 m/s
Solution:
Horizontal - 20×(cos 30°) = 5.18 m/s
Vertical - 20×(sin 30°) = 19.32 m/s
