What is the force constant of a spring which is stretcheda) 2 mm by a force of 4N,b) 4cm by a mass of 200g?.
1. What is the force constant of a spring which is stretcheda) 2 mm by a force of 4N,b) 4cm by a mass of 200g?.
Jawaban:
Untuk menghitung konstanta gaya suatu pegas, kita perlu menggunakan rumus:
k = F / x
Di mana:
k = konstanta gaya (dalam newton/meter)
F = gaya yang diberikan pada pegas (dalam newton)
x = perpanjangan pegas (dalam meter)
a) Dalam hal ini, gaya F = 4 N, dan perpanjangan pegas x = 2 mm = 2 x 10^-3 m. Jadi, konstanta gaya k = F / x = (4 N) / (2 x 10^-3 m) = 2000 N/m.
b) Dalam hal ini, massa m = 200 g = 0.2 kg. Selain itu, kita juga tahu bahwa gaya gravitasi pada bumi adalah F = m x g, di mana g adalah percepatan gravitasi (9.8 m/s^2). Jadi, gaya F = 0.2 kg x 9.8 m/s^2 = 1.96 N. Perpanjangan pegas x = 4 cm = 4 x 10^-2 m. Jadi, konstanta gaya k = F / x = (1.96 N) / (4 x 10^-2 m) = 49 N/m.
Penjelasan:
2. what is spring constant,if it is given force of 400n,and increace of 4cm length ....N/m
Jawaban:
F=400 N
4 cm = 0,04 m
[tex] \frac{400 \: n}{0.04} = 10.000 \: n \: per \: m[/tex]
3. When a 14.0 kg mass hangs from a spring that has a spring constant of 550 N/m, the spring has a length of 82 cm. Determine the length of the spring before any force is applied to it.
Penjelasan:
given :
m = 14.0kg
k = 550N/m
g= 10 m/s²
total length = 82cm = 0,82m
find:
the length of the spring before any force is applied to it, Lo =?
solutions:
F = m. g = 14.0 × 10 = 140N
F = k. ∆x
140 = 550.(0,82 -Lo)
14 = 55(0,82 -Lo)
14 = 45.1 - 55Lo
55Lo = 45.1 -14
= 31.1
Lo = 31.1/ 55
= 0.565m
= 56.5cm
4. what is the net force on an 800 kg airplane flying with a constant velocity of 160 km/hour north
Jawaban:
Punten kak, izin menjawab yaa..
Penjelasan:
According to Newton's Law (ll)
F = m.a
Because base on the question above that plane's velocity is constant, so delta v or we called "acceleration" is zero "0".
So, value of F on Newton's Law (ll) as follow:
F = 800 × 0 = 0
5. 5kg box is being lifted upward at a constant velocity of 2m/s by a force equal to the weight of the box.a) what is the power input of the force?b) How much work is done by the force in 4s?
Jawaban:
A. Force= 5kg×10kgm/s²(gravity)= 50 Newton
B. Work= 50 N×(4s×2m/s)= 400 Joule
Berikan penilaian terbaik ya
6. The aircraft landing gear consists of a spring- and hydraulically-loaded piston and cylinder D and the two pivoted links OB and CB. If the gear is moving along the runway at a constant speed with the wheel supporting a stabilized constant load of 24 kN, calculate the total force which the pin at A supports.
terjemahan :
roda pendaratan pesawat terdiri dari piston dan silinder D dan pegas yang dimuat secara hidrolik
dan dua pivot link OB dan CB .
jika Roda gigi bergerak di sepanjang runway
pada kecepatan konstan dengan roda mendukung beban konstan yang stabil sebesar
24 kN ,hitung total gaya yang di dukung pin pada A .
7. fisika kelas 10opsi ada di gambarThe launching mechanism of a toy gun consists of a spring of unknown spring constant. when the spring is compressed 0,1 m the gun, when fired vertically, is able to launch a 40 g projectile to a maximum height of 20.0 m above the position of the projectile before firing.
Jawabannya sudah saya lampirkan diatas.
8. a spring is given a force of 80 N and its length gwors by 2 cm. If the given force is 100 N, the growth in length is
80/2=100/x
80x=200
x=2,5 cm
9. derive the formula for work done by a constant force
Work is done when a force is applied, at least partially, in the direction of the displacement of the object. If that force is constant then the work done by the force is the dot product of the force with the displacement: W = F ⃗ ∙ d ⃗ .
10. . In a game of tug-of-war, both sides aren’t moving because each team have the same equal strength. What type of force is this? (10 Points) Balanced force Normal force Unbalanced force Spring Force
Jawaban: r u kelly from 6c
11. A box having a mass of 1.5 kg is accelerated across a table at 1.5 m/s2. the coefficient of friction on the box is 0.3. what is the force being applied to the box? if this force were applied by a spring, what would the spring constant have to be in order for the spring to be stretched to only 0.08 m while pulling the box?
Jawaban:
28.125 N/m
Penjelasan:
Untuk menghitung gaya yang diterapkan pada kotak, kita dapat menggunakan persamaan gaya = massa x akselerasi. Jadi, gaya yang diterapkan pada kotak adalah:
F = 1.5 kg x 1.5 m/s2 = 2.25 N
Untuk menghitung konstanta pegas yang diperlukan untuk menarik kotak sejauh 0,08 m, kita dapat menggunakan persamaan pegas: F = k x d (dimana F adalah gaya, k adalah konstanta pegas, dan d adalah jarak). Karena gaya yang diberikan adalah 2,25 N dan jarak adalah 0,08 m, konstanta pegas yang diperlukan adalah:
k = F / d = 2.25 N / 0.08 m = 28.125 N/m
12. A 10N force is needs to move an object with a constant velocity of 5m/s. What Power must be delivered to the object by the force?
P = W/t = (F s)/t = F x s/t = F x (v) = 10 x 5 = 50Wattp=w/t p=FV then p=10 x 5= 50 watt okay
13. the combination of an applied firce and friction force produces a constant
Jawaban:
[tex]\displaystyle I=21,6\,\text{kgm}^2\\\\\tau_f=3,6\,\text{Nm}\\\\n\approx52,4\,\text{rev}[/tex]
Penjelasan:
[tex]\displaystyle \tau =36\,\text{Nm}\\t_F=6\,\text{s}\\t_f=60\,\text{s}\\\omega_0=\omega_2=0\,\text{rad/s}\\\omega_1=10\,\text{rad/s}\\\\I=?\\\tau_f=?\\n=?\\\\\tau=I\alpha\\\tau=I\cdot\frac{\omega_1-\omega_0}{t_F}\\36=I\cdot\frac{10-0}{6}\\\boxed{\boxed{I=21,6\,\text{kgm}^2}}\\\\\omega_2=\omega_1-\alpha_2t_f\\\omega_2=\omega_1-\frac{\tau_f}{I}\cdot t_f\\0=10-\frac{\tau_f}{21,6}\cdot60\\0=1-\frac{\tau_f}{3,6}\\\boxed{\boxed{\tau_f=3,6\,\text{Nm}}}\\\\n=n_1+n_2\\n=\frac{\theta_1}{2\pi}+\frac{\theta_2}{2\pi}\\n=\frac{\omega_0t_F+\frac12\alpha t^2}{2\pi}+\frac{\omega_1t-\frac12\alpha_2 t^2}{2\pi}\\n=\frac{\omega_0t_F+\frac{\tau}{2I}t^2}{2\pi}+\frac{\omega_1t+\frac{\tau_f}{2I}t^2}{2\pi}\\n=\frac{0\cdot6+\frac{36}{2\cdot21,6}\cdot6^2}{2\pi}+\frac{10\cdot60-\frac{3,6}{2\cdot21,6}\cdot60^2}{2\pi}\\n=\frac{\frac{1}{2\cdot0,6}\cdot36}{2\pi}+\frac{600-\frac{1}{2\cdot6}\cdot3600}{2\pi}\\n=\frac{30}{2\pi}+\frac{300}{2\pi}\\n=\frac{330}{2\pi}\\\boxed{\boxed{n\approx52,4\,\text{rev}}}[/tex]
14. A spring with a constant of 80 N/m, is extended by 2 m. How much energy stored in the extended spring?
Energi Potensial pegas
Diketahui:
k = konstanta pegas = 80 N/m Δx = perpanjangan pegas = 2 mDitanya:
Ep = energi potensial pegas ? JawabanEp = ½ × k × Δx²
Ep = ½ × 80 × 2²
Ep = 40 × 4
Ep = 160 J ✔
15. 13.A spring has k=65 N/m. Draw a graph of force, F vs the spring elongation, r.Determine the work needed to stretch the spring from r=3.0 cm to r=6.5 cm,where r=0 refers to the spring outstretched length.
semoga bermanfaat dan membantu
16. In this discussion we wiil learn the effect of force on lenght increment of a spring. Pernyataan ini menyatakan tentang
In this materi, Discussion about effect of force on lenght increment of a spring.
17. Which of the following is a contact force?a. Electrostatic forcec. Gravitational forceb. Frictional forced. Magnetic force
Jawaban:
A. Electrostatic
Penjelasan:
Kayak ny menurut saya
18. Which of the following forces is acting on the book?O Frictional forceO Magnetic forceO Gravitational forceO Elastic spring force
Jawaban:
B. magnetic force
Penjelasan:
i hope this answer is true
19. A block of mass 15 kg executes simple harmonic motion under the restoring force of a spring. The amplitude and the time motion are 0.1 m and 3.14 s respectively. Find the maximum force exerted by the spring on the block !
Jawaban:
30 blok
Penjelasan:
20. what the example of spring force
when you click a pen or shockbaker in a motorcycle
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