Given that,
The force constant of a spring, k = 570.8 N/m
It is stretched 4.12 cm from the equilibrium.
To find,
The potential energy stored in the spring.
Solution,
The potential energy stored in the spring is given by the formula as follows :
[tex]E=\dfrac{1}{2}kx^2[/tex]
Put all the given values,
[tex]E=\dfrac{1}{2}\times 570.8 \times (4.12\times 10^{-2})^2\\\\E=0.484 J[/tex]
So, the required potential energy is 0.484 J.
A ball is projected at an angle of 30° above the horizontal with a speed of 35 m/s.What will be its approximate horizontal range across a level surface?
Answer:
24.89degree angle
Explanation:
Thats what i think because its my opinion it took me a while to answer your question but i wish you the best in luck with your degree angles
During a fireworks display, a shell is shot into the air at an angle of theta degrees above the horizontal. The fuse is timed to 8.6 seconds for the shell just as it reaches its highest point above the ground. If the horizontal displacement of the shell is 81.7 meters when it explodes, what is the angle theta of the initial velocity
Answer:
θ = 83.6º
Explanation:
Once the shell is in the air, the only influence acting on it (assuming that the resistance of the air is negligible) is gravity.Gravity acts on the shell accelerating it in the vertical direction with a=g=-9.8 m/s2 (taking the upward direction as the positive one).Since movements in perpendicular directions are independent each other, we conclude that in the horizontal direction, speed must be constant.So, we can find the speed in the horizontal direction (the projection of the velocity vector along the horizontal axis) applying simply the definition of average velocity, as follows:[tex]v_{avg} =\frac{\Delta x}{\Delta t} (1)[/tex]
Taking the launch site as the origin, and making the horizontal direction coincident with the x-axis (positive to the right), we know that the horizontal displacement is 81.7 m, and the time at the explosion, 8.6 s, so we can write the horizontal component of the velocity vector (its projection along the x-axis) as follows:[tex]v_{x} = \frac{81.7m}{8.6s} = v_{o} * cos \theta = 9.5 m/s (2)[/tex]
In the vertical direction, since we know that the shell will explode when it reaches to the maximum height, at this point just before exploding, the vertical component of the velocity is just zero, even though it continues being accelerated downward.Applying the definition of acceleration, replacing a by the value of g, we can write the following expression:[tex]v_{fy} = v_{oy} - g*t = v_{o} * sin \theta -g*t = 0 (3)[/tex]
⇒ v₀* sin θ = g*t = 9.8m/s2* 8.6s = 84.3 m/s (4)
Dividing (4) by (2), we can find tg θ (the angle of the initial velocity with the x-axis), as follows:[tex]\frac{v_{o}* sin \theta}{v_{o}* cos \theta} = tg \theta= \frac{84.3m/s}{9.5m/s} = 8.87 (5)[/tex]
⇒ θ = tg⁻¹ (8.87) = 83.6º
A 5 kilogram block is dropped from a height of 3 meters and falls straight to the ground. What is the work done by the force of gravity?
Answer:
Workdone = 147Nm
Explanation:
Given the following data;
Mass = 5kg
Height = 3m
We know that acceleration due to gravity is 9.8m/s²
First of all, we would find the force applied;
[tex] Force = mass * acceleration [/tex]
Force = 5*9.8
Force = 49N
Now, to find the workdone;
[tex] Workdone = force * distance [/tex]
Substituting into the equation, we have;
Workdone = 49 * 3
Workdone = 147Nm
Find the value of x help asap
Answer:
46 degrees
Explanation:
90+44+x=180
44+x=90
x=46
Answer:well isn't it 90 like it said on the image?
Explanation?????
a small rocket is launched vertically upward from the edge of a cliff 400 ft off the ground at the speed of 200 ft/s. its height above the ground is given by the function h(t)
Question:
A small rocket is launched vertically upward from the edge of a cliff 80 ft off the ground at a speed of 69 ft/s. Its height (in feet) above the ground is given by [tex]h(t) = - 25t^2 + 200t + 400[/tex] where t represents time measured in seconds.
Assuming the rocket is launched at t=0, what is an appropriate domain for h?
Answer:
[tex]0 \le t \le 9.66[/tex]
Explanation:
Given
[tex]h(t) = - 25t^2 + 200t + 400[/tex]
Required
Determine the domain of h
The initial value of t is 0.
i.e.
[tex]t = 0[/tex] --- This is given in the question
To determine the final value, we simply solve for t in [tex]h(t) = - 25t^2 + 200t + 400[/tex]
Rewrite the expression:
[tex]- 25t^2 + 200t + 400=0[/tex]
Divide through by -25
[tex]\frac{-25t^2}{-25} + \frac{200t}{-25} + \frac{400}{25}=\frac{0}{-25}[/tex]
[tex]t^2 - 8t - 16 = 0[/tex]
Using quadratic formula:
[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]
Where
[tex]a= 1[/tex] [tex]b = -8[/tex] and [tex]c = -16[/tex]
The expression becomes:
[tex]t = \frac{-(-8)\±\sqrt{(-8)^2 - 4*1*(-16)}}{2*1}[/tex]
[tex]t = \frac{8\±\sqrt{64+ 64}}{2}[/tex]
[tex]t = \frac{8\±\sqrt{128}}{2}[/tex]
[tex]\sqrt{128} = 11.31[/tex]
So, we have:
[tex]t = \frac{8\±11.31}{2}[/tex]
Split
[tex]t = \frac{8+11.31}{2}\ or\ t = \frac{8-11.31}{2}[/tex]
[tex]t = \frac{19.31}{2}\ or\ t = \frac{-3.31}{2}[/tex]
[tex]t = 9.66\ or\ t = -1.66[/tex]
But time can't be negative.
So:
[tex]t = 9.66[/tex]
Hence, the domain is:
[tex]0 \le t \le 9.66[/tex]
An object travels at 3.06 m distance with uniform acceleration in this interval it's velocity increases by 0.18 and it's average velocity over the interval was 0.34 m/s what is the acceleration?
Given that,
Distance travelled, d = 3.06 m
Final velocity, v = 0.18 m/s
Average velocity of the object, V = 0.34 m/s
To find,
The acceleration of the object.
Solution,
Let t is the time in which it covers distance d. Average velocity is given by distance covered divided by time.
[tex]V=\dfrac{d}{t}\\\\t=\dfrac{d}{V}\\\\t=\dfrac{3.06}{0.34}\\\\t=9\ s[/tex]
Let a is the acceleration of the object. Using equation of motion to find it :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0.18-0}{9}\\\\a=0.02\ m/s^2[/tex]
So, the acceleration of the object is 0.02 m/s².
A man is picked up by a tornado and dropped a quarter mile away from his home. Even though authorities estimate that his body was dropped at 30 mph he suffered no major injuries. What factors contributed to the fact that his body was able to withstand this drop?
Answer:
the factor that contributed to the fact that he was able to survive this fall is due to the fact that his body is made up of bone which are strong and his skeleton also protected all his vitals internal organs. His body went limp after he was knocked in the head when he fell, all of these made room allowing his muscles to relax and this gave his bones the opportunity to absorb the impact of the fall
If a man was dropped by the tornado from a distance of a quarter away from home. As per the authorities, he was dropped least at a speed of 30 mph. The man suffered no main injuries and survived.
The factors that would have contributed to the man's body able to withstand the drop were the air current that were so strong that he may have landed in a safe place. The fact that he fell from lesser height or the man might have been fat.Learn more about the up-by a tornado and dropped a quarter-mile away.
brainly.com/question/1492400.
The graph represents velocity over time.
A graph with horizontal axis time (seconds) and vertical axis velocity (meters per second). A line runs from 0 seconds 2 meters to 100 seconds 22 meters.
What is the acceleration?
–0.2 m/s2 I did it on edge 2020
Answer:
the answer is C
Explanation:
i got it wrong the first time because the answer above says -0.2 but its actually 0.2
i hope this helps
A Soccer player kicks a soccer ball rolling on the floor with a force of
1,600 N, for.022 seconds. What is the impulse or Ap of the soccer
ball?
Answer:
35.2Ns
Explanation:
The change in momentum of a body is also its impulse;
Given:
Force of kicking = 1600N
Time of action = 0.022 s
Unknown:
Impulse = ?
Solution:
The impulse on the body is;
Impulse = Force x time
Now;
Impulse = 1600 x 0.022 = 35.2Ns
How many significant figures are there in the measurement (4.07×10^17m)
Answer and I will give you brainiliest
please heeeelp