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2 years ago

A 20 kg weight is suspended from a string. What is the tension on the string

2 answers:

7 0

The weight is in equilibrium, so the two forces acting on it (tension T pulling upward, and its own weight W pulling it downward) cancel each other. By Newton's second law,

∑ F = T - W = 0

so that

T = W

Calculate the object's weight:

W = (20 kg) (9.8 m/s²) = 196 N

So the tension is T = 196 N as well.

JulsSmile [24]2 years ago

3 0

assume that tension in string a is Ta and in string b is Tb

now taking the component of tensions

Tasin30∘=Tbsin30∘...(1)

it means Ta=Tb

Tacos30∘+Tbcos30∘=mg...(2)

⇒2Tacos30∘=20×10⇒Ta=Tb=3200

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2 years ago

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Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

Initial Velocity of ball V_i,b = 0

Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p) ...2

The distance that the platform moves d:

d = V_f,p *t ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p) .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

The platform moves 0.05312 m to the opposite to which the ball is thrown.

5 0

2 years ago

The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decre

Andrew [12]

Answer:

0.34

Explanation:

2.5 Mg = 2500 kg

The change in speed from 100 km/h to 40 km/h is

$\Delta v = 100 - 40 = 60 km/h = 60 * 1000 / (60 * 60) = 16.67 m/s$

The deceleration caused by friction force is the change in speed per unit of time

$a = \Delta v / \Delta t = 16.67 / 5 = 3.33 m/s^2$

Using Newton 2nd law we can calculate the friction force that caused this deceleration:

F = ma = 2500 * 3.33 = 8333 N

Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

$F = mg\mu$

$8333 = 2500*9.8\mu$

$\mu = \frac{8333}{2500 * 9.8} = 0.34$

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3 years ago