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mixas84 [53]
3 years ago
15

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

ity of the bullet?
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

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The final acceleration becomes (1/3) of the initial acceleration.

Explanation:

The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

F=ma

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a = acceleration

According to given condition, if the mass of a  sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

a=\dfrac{F}{m}

Let a' is the final acceleration,

a'=\dfrac{F}{m'}

m' = 3m

a'=\dfrac{F}{3m}

a'=\dfrac{1}{3}\times \dfrac{F}{m}

a'=\dfrac{1}{3}\times a

So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.

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3 years ago
A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acce
Vera_Pavlovna [14]

Answer:

D) 15s

Explanation:

let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.

the period of a pendulum is given by:

T = 2π√(L/g)

so on earth:

Te = 2π√(L/g1)

     =  6s

on the moon;

Tm = 2π√(L/g2)

since g2 = 1/6 g1 then:

Tm = 2π√(L/(1/6×g1))

      = √(6)×2π√(L/(g1))

and 2π√(L/(g1)) = Te = 6s

Tm = (√(6))×6 = 14.7s ≈ 15s

Therefore, the period of the block-spring system on the moon is 15s.

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A hybrid is
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The correct answer is heterozygous.
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4 years ago
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Which of the following is a true statement for a child's toy spinning in a circle at constant speed?
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The answer is C.

Explanation:

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An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o
klio [65]

Answer:

3.58\:\mathrm{s}

Explanation:

We can use the kinematics equation \Delta y=v_it+\frac{1}{2}at^2 to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}

Now we can substitute values in our kinematics equation:

  • \Delta y=-100
  • a=-9.8\:\mathrm{m/s^2} (acceleration due to gravity)
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  • Solving for t

-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}

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3 years ago
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