Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question
Answer:
well... when the horse stops/rests, or if it is blocked by a surface or anything of solid background.
Explanation:
If it is going up a hill or slope and it just starts to move that would also be considered the smallest amount of acceleration this can go for many things when it just starts to move. but I would go for when it rests amounting to your fitting of the question.
Answer:
W = 30 J
Explanation:
given,
Work done = 10 J
Stretch of spring, x = 0.1 m
We know,
dW = F .dx
we know, F = k x
![W = k[\dfrac{x^2}{2}]_0^{0.1}](https://tex.z-dn.net/?f=W%20%3D%20k%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_0%5E%7B0.1%7D)
k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.
![W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx](https://tex.z-dn.net/?f=W%20%3D%202000%20%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0.1%7D%5E%7B0.2%7D%20dx)
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
Answer:
Spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.
Explanation:
Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in <u>elastic region</u>, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.
Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.
Answer:
15 cm
Explanation:
Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:
Gaya 1 (F₁) = 225 N
Jarak terpisah 1 (d) = 5 cm
Gaya 2 (F₂) = 25 N
Jarak terpisah 2 (d₂) =?
Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:
F = Kq₁q₂ / d²
Perbanyak silang
Fd² = Kq₁q₂
Menjaga Kq₁q₂ konstan, kita memiliki:
F₁d₁² = F₂d₂²
Dengan rumus di atas maka diperoleh jarak sebagai berikut:
Gaya 1 (F₁) = 225 N
Jarak terpisah 1 (d) = 5 cm
Gaya 2 (F₂) = 25 N
Jarak terpisah 2 (d₂) =?
F₁d₁² = F₂d₂²
225 × 5² = 25 × d₂²
225 × 25 = 25 × d₂²
5625 = 25 × d₂²
Bagilah kedua sisinya dengan 25
d₂² = 5625/25
d₂² = 225
Hitung akar kuadrat dari kedua sisi
d₂ = √225
d₂ = 15 cm
Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N
