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Vitek1552 [10]
3 years ago
5

What do odometer and speedometer read ?

Physics
1 answer:
maksim [4K]3 years ago
6 0
Odometer: tells you the distance traveled by vehicle since it was new (or when last reset)

Speedometer: tells you the velocity of the vehicle at that moment.
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A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

8 0
2 years ago
Read 2 more answers
Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before col
Akimi4 [234]

Answer:

va = 4.79 m/s

vb = 1.29 m/s

Explanation:

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂

22.75 = 3v₁ + 6.5v₂

For an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²

79.625 = 3v₁² + 6.5v₂²

Two equations, two variables.  Solve with substitution:

22.75 = 3v₁ + 6.5v₂

22.75 − 3v₁ = 6.5v₂

v₂ = (22.75 − 3v₁) / 6.5

79.625 = 3v₁² + 6.5v₂²

79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²

79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5

517.5625 = 19.5v₁² + (22.75 − 3v₁)²

517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²

0 = 28.5v₁² − 136.5v₁

0 = v₁ (28.5v₁ − 136.5)

v₁ = 0 or 4.79

We know v₁ isn't 0, so v₁ = 4.79 m/s.

Solving for v₂:

v₂ = (22.75 − 3v₁) / 6.5

v₂ = 1.29 m/s

8 0
2 years ago
Red-orange sunsets and blue skies are a result of what type of light-matter interaction?
fomenos
B.        polarization
 i hope that is correct

8 0
2 years ago
Read 2 more answers
1. A man is running on the straight road with the uniform velocity
Rufina [12.5K]
None he is undergoing constant uniform velocity. Acceleration is the rate of change of velocity
8 0
3 years ago
Read 2 more answers
(30 Points) A piston–cylinder arrangement containing Nitrogen (N2), initially at 4 bar and 438 K, undergoes an expansion to a fi
BartSMP [9]

Answer:

A. Final pressure P2

P2/P1 = (T2/T1)^n/n-1

P1 = 4bar

T1 = 438K

T2 = 300K

Polytropic index, n, = 1.3

P2 = 4 (300/438)^1.3/1.3-1

P2 = 4 (300/438)^4.333

P2 = 4 * 0.19400

P2 = 0.776bar.

B. The work done is;

W = mR/ n-1 (T1 -T2)

Where, R = 0.1889kJ/kg.K, m = 1

W = 1 * 0.1889/ 1.3-1 * (438-300)

W = 86.89kJ/kg.

C. The heat transfer, Q

Q = W + ΔU

Q = W + mCv(T2-T1), where Cv of nitrogen is 0.743kj/kgk

Q = 86.89 + 1 * 0.743 (300-438)

Q = 86.89 + (-102.534)

Q = -15.644kJ/K

Q = 15.64kJ/K

3 0
3 years ago
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