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mestny [16]
2 years ago
15

2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/

s east. Calculate the final speed and direction of the magnetic car (coupled) system?
Physics
1 answer:
Afina-wow [57]2 years ago
4 0

Answer:

2.44 m/s due East

Explanation:

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

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

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Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i
spayn [35]

Answer:

v = 15.56 m/s

v = 56 km/h

Explanation:

When coefficient of friction is approximately zero then we have

F_ncos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

tan\theta = \frac{v^2}{Rg}

here we know that

v = 40 km/h = 11.11 m/s

R = 30 m

tan\theta = \frac{11.11^2}{30\times 9.81}

\theta = 22.75 degree

now when friction coefficient is 0.30 then we have

F_n cos\theta = mg + F_f sin\theta

F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}

now we have

v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}

v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}

v = 15.56 m/s

v = 56 km/h

3 0
2 years ago
What type of stress is caused by two tectonic plates sliding past one another
prisoha [69]
Extensional stress. is your answer.
5 0
3 years ago
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Which type of constraint typically requires a longer time to change?
Mars2501 [29]
Structural constraint is the answer :)
3 0
3 years ago
From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0
2 years ago
Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, p
Studentka2010 [4]

Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

8 0
3 years ago
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