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4 years ago

The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n

1 answer:

5 0

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

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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3 years ago

What is the difference between speed and velocity?

mixer [17]

Speed is the rate of change of distance with time while velocity is the rate of change of displacement with time.
Speed is a scalar quantity while velocity is a vector quantity.
Speed cannot be negative but velocity can be negative.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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2 years ago

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Answer:

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Explanation:

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3 years ago

A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the

Norma-Jean [14]

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

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Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

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