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vichka [17]
2 years ago
12

An 8 g bullet leaves the muzzle of a rifle with

Physics
1 answer:
Elena-2011 [213]2 years ago
4 0

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

V^{2}=V_{o}^{2} + 2ad (1)

F=ma (2)

Where:

V=611.9 m/s is the bullet's final speed (when it leaves the muzzle)

V_{o}=0 is the bullet's initial speed (at rest)

a is the bullet's acceleration

d=0.8 m is the distance traveled by the bullet before leaving the muzzle

F is the force

m=8 g \frac{1 kg}{1000 g}=0.008 kg is the mass of the bullet

Knowing this, let's begin by isolating a from (1):

a=\frac{V^{2}}{2d} (3)

a=\frac{(611.9 m/s)^{2}}{2(0.8 m)} (4)

a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2} (5)

Substituting (5) in (2):

F=(0.008 kg)(2.34(10)^{5} m/s^{2}) (6)

Finally:

F=1872 N

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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u
Readme [11.4K]

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

d_1 = v*t_1

d_1 = 10*20 = 200 m

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

d_2 = v*t = 20*20 = 400 m

so the net displacement is given as

\vec d = \vec d_2 - \vec d_1

\vec d = 400 - 200 = 200 m

so it will be displaced by total displacement 200 m

8 0
3 years ago
1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir
Inga [223]

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

  • [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
  • [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

8 0
1 year ago
A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at
Jobisdone [24]

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = <em>360 km-m/s </em>

The direction of the change is <em>up /\ </em>.

8 0
3 years ago
Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
nignag [31]
<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

3 0
3 years ago
Find :
Pepsi [2]

Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

a) h =  v₀t + ½gt² = 40(4) +  ½(-10)4² = 80 m

or

v² = u² + 2as

h = (0² - 40²) / 2(-10) = 80 m

8 0
2 years ago
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