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8_murik_8 [283]
3 years ago
6

A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur

face of the ramp, and the ramp itself rises at 25.0° above the horizontal. Determine the tension in the cable?
Physics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force:  at angle of 31.0° above the surface of the ramp

Calculated of the Weight  of the car (W)

W = m*g   m: mass   g:acceleration due to gravity

W =   1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx =  11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T =  4680.07 / cos 25.0°

T = 5163.89 N

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Body waves are of two types:

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These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

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What is the frequency of a wave with a wavelength of 12 meters and a velocity of 4 m/s?
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Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa, 518 oC, and mass flow rate of 5,322 kg/h and exits the system at 96
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Explanation:

Pressure at inlet P_{1} = 1 × 10^{6} Pa

Temperature at inlet T_{1} = 518 ° c = 791 K

Mass flow rate = \frac{5322}{60} \frac{kg}{sec} = 88.7

Gas constant for carbon die oxide is R = 189 \frac{J}{kg k}

Mass flow rate inside the nozzle is given by the formula = \frac{P_{1} }{R T_{1} } × A_{1} × V_{1}

⇒ P_{1} = = 1 × 10^{6} Pa

⇒ RT_{1} = 791 × 189 = 149499 \frac{J}{kg}

⇒ A_{1} = 0.0037 m^{2}

Put all the above values in above formula we get,

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The acceleration that the same force will provide if both masses are tied together is; 6.0 m/s².

<h3>How to find the Acceleration?</h3>

We are given;

Force; F = 5 N

Acceleration of the first mass, a₁ = 8.0 m/s²

Acceleration of the second mass, a₂ = 24 m/s²

Formula for force is;

F = ma

Let us find both masses; m₁ and m₂.

m₁ = F/a₁

m₂ = F/a₂

Thus;

m₁ = 5/8 kg

m₂ = 5/24 kg

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Thus, acceleration if they are both tied together is;

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a = 6.0 m/s².

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