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Maurinko [17]
4 years ago
9

A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment

um in the northward direction?
A. +15,848 kg-m/s
B. +33,987 kg-m/s
C. +37,500 kg-m/s
D. +339,865 kg-m/s
Physics
1 answer:
Marianna [84]4 years ago
8 0

C!!!+37,500 is the answer

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Suppose two equal mass cars traveling with equal speeds in opposite directions inelasticcollide head on and stick together. What
Annette [7]

A collision which is inelastic is one in which the internal energy changes.

All the kinetic energy is dissipated

Reasons:

The given parameters are;

The mass of one car = The mass of the second car in the collision = m

The type of collision = Inelastic collision

The speed of each of the cars in the collision = v and -v

Solution:

In an inelastic collision, we have;

m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = (m_1 + m_2) \cdot v_f

Therefore, we get;

m \cdot v_{i} -m \cdot v_{i} = 0= 2 \cdot m \cdot v_f

Which gives;

v_f = 0

The collision is an example of a perfectly inelastic collision, and the final velocity after the collision is zero.

The change in the energy is \Delta K.E. = K.E._{final} - K.E._{initial}

K.E._{initial} = 0.5·m·v² + 0.5·m·v² = m·v²

ΔK.E. = 0.5·(2·m)×0 - 0.5·m·v² + 0.5·m·v² = -m·v²

ΔK.E. = -m·v²

The negative sign stands for energy given out, which gives;

The energy dissipated = m·v² (The total initial kinetic energy)

Therefore, <u>all the kinetic energy is dissipated</u>

Learn more here:

brainly.com/question/19106769

5 0
3 years ago
On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed
madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

K_{1}  = K_{2} + W_{loss}

Where:

K_{1}, K_{2} - Initial and final kinetic energy, measured in joules.

W_{loss} - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

\frac{1}{2}\cdot m \cdot v_{1}^{2}  = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s

Where:

v_{1}, v_{2} - Initial and final speed of the skater, measured in meters per second.

m - Mass of the skater, measured in kilograms.

\Delta s - Length of the rough patch, measured in meters.

f - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

f = r \cdot m \cdot g

Where:

r - Ratio of friction force to weight, dimensionless.

g - Gravitational constant, measured in meters per square second.

Then,

\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s

The equation is simplified algebraically and patch length is cleared afterwards:

\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s

\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }

Given that v_{1} = 5\,\frac{m}{s}, v_{2} = 2.5\,\frac{m}{s}, r = 0.22 and g = 9.807 \,\frac{m}{s^{2}}, the length of the rough patch is:

\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta s = 4.345\,m

The length of the rough patch is 4.345 meters.

6 0
3 years ago
Momentum is a scalar quantity.<br>True or false​
Rudiy27
The answer is false
5 0
3 years ago
Read 2 more answers
Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the volume flow rate of the fluid in m^3/
fenix001 [56]

Answer:

Volume flow rate = 1.41 \times 10^{-3}m^3/s

Explanation:

The volume flow rate through a channel can be gotten by multiplying its area and its velocity.

The channel under consideration is a circular channel. Hence, the cross-sectional area can be calculated by using the relation for calculating the area of a circle

Cross-sectional area = \pi \times d^{2}/4

A = \pi \times (3\times 10^-2)^2  /4 = 7.07 \times 10 ^-4 m^2

volume flow rate= A =  7.07 \times 10 ^-4 m^2 X 2.0 = 1.41 \times 10^{-3}m^3/s

Volume flow rate = 1.41 \times 10^{-3}m^3/s

7 0
4 years ago
A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and t
vladimir2022 [97]

This question is incomplete, the complete question is;

Flag

A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

8 0
3 years ago
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