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grigory [225]
2 years ago
14

A truck of mass mT = 2000 kg is going north on Guadalupe with a speed of 4 m/s. The truck is struck by an eastbound car (mass of

1000 kg) traveling on 24th street. From the skidmarks after the collision, it is deduced that the mangled wreckage, which sticks together after the collision was skidding at angle 20◦ with respect to the initial direction of the car. What was the initial speed of the car?
Physics
1 answer:
pogonyaev2 years ago
5 0

Answer:21.97 m/s

Explanation:

Mass of truck m_T=2000 kg

Velocity of truck v_c=4 m/s

Mass of car m_c=1000 kg

let v_c be the car velocity and u be the velocity of combined system after collision at angle of 20^{\circ}

Conserving momentum in east and North direction Respectively

In east Direction

m_c\cdot v_c=(m_c+m_T)u\cos 20-------1

In North direction

m_T\cdot v_T=(m_c+m_T)u\sin 20---------2

Divide 1 and 2 we get

\frac{m_T\cdot v_T}{m_c\cdot v_c}=\frac{(m_c+m_T)u\sin 20}{(m_c+m_T)u\cos 20}

v_c=\frac{m_T\cdot v_T}{m_c\cdot \tan 20}

v_c=\frac{2000}{1000}\times \frac{4}{\tan 20}

v_c=21.97 m/s

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KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

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180 = 200 sin 2πft

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sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

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360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

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2 years ago
A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme
GaryK [48]

Given that,

Distance in south-west direction = 250 km

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cos ∅ = base/hypotenuse

base= hyp * cos ∅

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East component = 125 km

8 0
2 years ago
Read 2 more answers
Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder
Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

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a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

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a is linear acceleration

α is angular acceleration

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α=a/r

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α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

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v=u+at

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7 0
2 years ago
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

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Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

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F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

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Drag force = 14422.2 N acting opposite to the motion

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TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

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7 0
2 years ago
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Arada [10]

Answer:

an absorber of x-ray

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3 0
2 years ago
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