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Vera_Pavlovna [14]
2 years ago
10

A small car of mass 1200 kg traveling east at 60m/s collides at an intersection with a truck of mass 3000 kg that is traveling d

ue north at 40 m/s. The two vehicles are locked together. What is the velocity of the combined wreckage
Physics
2 answers:
lakkis [162]2 years ago
5 0

Answer:

11.4m/s I think mb not 100% lol

Explanation:

P=mv

P1 = 1200×60 = 72000

P2 = 3000×40 = 120000

120000+(-72000) = 48000

v=P÷m

m=1200+3000 = 4200

48000÷4200 = 11.4m/s

erik [133]2 years ago
5 0

v2f= 2⋅m1 (m2+m1) v1i + (m2−m1)(m2+m1) v2i v 2

meaning that

2 x 1200 ( 3000 + 1200) 60 + (3000-1200)(3000+1200) 40f =

2400 (4200) 60 + (1800) (4200) 40f =

10080000 (60) +  75600000 (40)

604800000 + 3024000000

3628800000? this is what i got

f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i

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zmey [24]

Compound machines have two or more simple machines

3 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
3. How can we determine the volume of a regular object?
JulsSmile [24]
Length•Width•Height is the answer
4 0
3 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Big B
Komok [63]

The distance covered by an object accelerating from rest is

D = (1/2) · (acceleration) · (time)² .

In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.

D = (1/2) · (9.8 m/s²) · (1.67 s)²

D = (4.9 m/s²) · (2.789 s²)

D =  13.67 meters

6 0
3 years ago
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