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marshall27 [118]

2 years ago

13

A 10kg toy truck has a 5kg toy car at rest. If the toy truck was moving at 3 m/s before the collision and carries that car with

it, what is the
Final velocity of the car and truck.
A. 15 m/ s
B. 30 m/ s
C 2 m/ s
D. 18 m/ sc free

2 answers:

dolphi86 [110]2 years ago

7 0

Answer:

Let m1 = mass of big toy car=10kg

m2= mass of small toy car= 5kg

U1= initial velocity of big toy car= 3m/s

U2= initial velocity of small toy car=0

Since the big toy car moved the small one after the collision, their final velocity will be the same.

m1u1 + m2u2= (m1+m2)v

10(3)+(5)(0)=(10+5)v

30=15v

Divide both sides by 15

V=2

The final velocity is 2m/s

Explanation:

klasskru [66]2 years ago

5 0

Answer:

$v_f=2\:\mathrm{m/s}$

Explanation:

From the Law of Conservation of Momentum, momentum is conserved. Therefore, we can set up the following equation:

$m_1v_1+m_2v_1=m_fv_f$

Since the 5kg toy car was initially and rest, $m_2v_2=0$ .

Therefore, plugging in our values, we have:

$10\cdot 3=(10+5)v_f,\\v_f=\frac{30}{15},\\v_f=\fbox{$2\:\mathrm{m/s}$}$ .

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frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

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zhuklara [117]

Answer:

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Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

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now we have

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