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Elden [556K]
3 years ago
14

An unstretched ideal spring hangs vertically from a fixed support. A 0.4 kg object is then attached to the lower end of the spri

ng. The object is pulled down to a distance of 0.35 m below the unstretched position and released from rest at time t= 0. A graph of the subsequent vertical position y of the lower end of the spring as a function of t is given above, where y= 0 when the spring was initially unstretched. At which time is the upward velocity of the object the greatest?
Physics
1 answer:
musickatia [10]3 years ago
8 0

Answer:

The correct answer will be "0.25 sec".

Explanation:

The graph of the given question is attached below.

According to the graph of the question,

Time,

T = 1 sec

For the upward velocity,

⇒ t = \frac{T}{4}

By putting the value, we get

⇒    =\frac{1}{4}

⇒    =0.25 \ sec

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An angry rhino with a mass of 2600 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distr
worty [1.4K]

Answer:

vf₂ = 15.79 m/s

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf  Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁ = 2600 kg : mass of the rhino

m₂= 0.18 kg : mass of the ball

v₀₁ = 3.70 m/s  : initial velocity of the rhino

v₀₂= - 8.39 m/s, initial velocity of the ball

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

(2600)*(3.7) + (0.18)*(- 8.39) = (2600)*vf₁ +(0.18)*vf₂

9620 -1.5102 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e = \frac{v_{f2}-v_{f1} }{v_{o1}-v_{o2}}

1*( v₀₁- v₀₂ )  = (vf₂ -vf₁)

( 3.7 -( -8.39 )  = (vf₂ -vf₁)

12.09 =  (vf₂ -vf₁)

vf₂ = vf₁ + 12.09 Equation (2)

We replace Equation (2) in the Equation (1)

9618.4898 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)* (vf₁ + 12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + (0.18)(12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + 2.1762

9618.4898 -2.1762 = (2600.18)*vf₁

9616.3136 = (2600.18)*vf₁  

vf₁ = (9616.3136) / (2600.18)

vf₁ = 3.698 m/s : Final velocity of the rhino

We replace vf₁ = 3.698 m/s in the Equation (2)

vf₂ = vf₁ + 12.09

vf₂ = 3.698 + 12.09

vf₂ = 15.79 m/s : Final velocity of the ball

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A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh
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Answer:

Explanation:

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