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NemiM [27]
3 years ago
7

In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a

t 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they couple (attach together) and roll together with a speed of __________. Ignore resistive forces.a. 0.5 m/s
b. 0.33 m/s
c. 0.67 m/s
d. 1.0 m/s
e. none of these
Physics
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

u_1,u_2 = initial velocity before collision

v_1,v_2 = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

m_1=m_2=M

Initial velocity of m_1=u_1=2 m/s

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

Final velocity of m_1=v_1=v (same direction )

Final velocity of m_2=v_2=v  (same direction)

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

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Explanation:

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An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes
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<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

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6 0
2 years ago
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
Can you please tell me what this is
Anna [14]

Answer:

200000 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE => is the kinetic energy.

m =>is the mass of the object

V => it the velocity of the object.

With the above formula, we can obtain the kinetic energy of the roller coaster as follow:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 1000 × 20²

KE = 500 × 400

KE = 200000 J

Therefore, the kinetic energy of the roller coaster is 200000 J.

4 0
3 years ago
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