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1 year ago

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before

Physics

1 answer:

RideAnS [48]1 year ago

7 0

The velocity of pluck 1 is 12 m/s west.

<h3>What is the conservation of momentum?</h3>

The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.

Now given that;

m1u1 + m2u2 = m1v1 + m2v2

(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)

1.5 - 1.2 = 0.1v + 1.5

0.3 - 1.5 = 0.1v

v = -1.2/0.1

v = - 12 m/s

Hence, the velocity of pluck 1 is 12 m/s west.

Learn more about linear momentum:brainly.com/question/27988315

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3 years ago

Two blocks, joined by a string, have masses of 6.0 and 9.0 kg. They rest on a frictionless, horizontal surface. A second string,

Lynna [10]

Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting on both masses, we can write both equations as follows:

Fy = N- (m₁+m₂)*g = 0 (as there is no movement in the vertical direction)

Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

4 0

3 years ago

A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is

Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

Force applied = 100 N
Distance covered = 5 metres
Time = 10 seconds

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

$\bf \boxed{Work = Force \times displacement}$

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

$\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}$

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

5 0

3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

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 $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago