You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N

5 0

3 years ago

If the only forces acting on a 2.0kg mass are F1 = (3i-8j)N and F2 = (5i+3j)N, what is the magnitude of the acceleration of the

belka [17]

Answer: 4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

3 0

3 years ago

A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu

neonofarm [45]

Answer:

$f = ma \\$

$w \: = \frac{f}{g} \\$

Explanation:

$f \: = ma \\ f - mg = ma \\ f = ma + mg \\ f = 40 \times 2 + 40 \times 10 \\ f = 480$

$w = \frac{f}{g} \\w = \frac{480}{10} \\ w = 48kg$

5 0

3 years ago