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larisa86 [58]
3 years ago
15

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

e bullet has a mass m, a speed v before the collision with the target, and a speed (0.516)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.413)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)
Physics
1 answer:
NikAS [45]3 years ago
7 0

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

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LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
  • u = 48/16
  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
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Hence, The magnification of the ornament is 0.25

8 0
2 years ago
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scoray [572]
Newton's 2nd law of motion:

                       Force  =  (mass) x (acceleration)

                                   =  (1,127 kg) x (6 m/s² forward)

                                   =  (1,127 x 6)  newtons forward

                                   =    6,762 newtons forward
______________________________

             Momentum  =  (mass) x (speed)

                                 =   (69 kg) x (6 m/s)

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3 0
3 years ago
A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu
bija089 [108]

Answer:

<em>J=36221 Kg.m/s</em>

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed v_1 and changes it to v_2. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

p=mv

The initial and final momentums are, respectively

p_1=mv_1,\ p_2=mv_2

The change of momentum is

\Delta p=p_2-p_1=m(v_2-v_1)

It is numerically equal to the Impulse J

J=\Delta p

J=m(v_2-v_1)

We are given

m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s

The impulse the car experiences during that time is

J=2498(2.6-17.1)=2498(-14.5)

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
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