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marusya05 [52]
3 years ago
5

A tow truck is pulling a car out of a ditch by means of a steel cable (y = 2.0 x 1011 n/m2) that is 9.55 m long and has a radius

of 0.437 cm. when the car just begins to move, the tension in the cable is 803 n. how much has the cable stretched?
Physics
1 answer:
Anna11 [10]3 years ago
4 0
If you know the real modulus of the cable (Y), the length, and the area (based on the radius), you can compute the spring constant, k = AE/L. Then, if you know the force used, you can compute the displacement, using F = kd, or d = F / k = FL/(AE). Our answer should work out to units of length. So, 
d = 803 N * 9.06 m / [pi*(0.574 cm)^2 * 2.0 x 10^11 N/m^2] 
d = 3.5 x 10^-8 Nm^3 / (cm^2 * N) 
d = 3.5 x 10^-8 m^3 / cm^2 * (100 cm / 1 m)^2 
d = 3.5 x 10^-4 m
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Some engineers have suggested that we can simulate gravity in outer space by having a circular rotating space station where pers
Thepotemich [5.8K]

Answer:

Centrifugal force

Explanation:

In Space, It is in a way possible create an " artificial gravity" just by spinning the aircraft or space station. On spinning the space station the inhabitants feel an outward force or the centrifugal force, this outward force when equivalent to gravitational force is able to stimulate gravity.

8 0
3 years ago
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grainsare there in the ball
Alina [70]

Answer:

the number of grains in the ball is 274,848

Explanation:

Given that;

diameter = 0.5 mm

so radius r = 0.25 mm

first we determine the volume of the ball using the following equation;

V = 4/3×πr³

we substitute

V = 4/3×π(0.25)³

V =  0.06544 mm³

Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³

so;

Number of grains N = 0.06544 × 4,200,000

N = 274,848 grains

Therefore, the number of grains in the ball is 274,848

5 0
3 years ago
4. A car of mass 2000 kg is traveling at 45 m/s when the driver spots a policeman anead. i ne univer apprivo
MariettaO [177]

Answer:

The change in the Kinetic Energy of the car, E = 1449000 joules

Explanation:

Given,

The mass of the car, m = 2000 Kg

The speed of the car, v = 45 m/s

The brake applied on the car for a duration, t = 3 s

The average force applied by the brake, F = 1.4 x 10⁴ N

The kinetic energy of the body is given by the relation,

                                     K.E = 1/2 mv²

The initial kinetic energy of the car,

                                   K.E = 0.5 x 2000 x 45

                                          = 2025000 J

The force applied by the brakes

                                   F = m x a

Therefore, the deceleration of the car

                                     a = F / m

                                        = 1.4 x 10⁴ / 2000

                                       = 7 m/s²

Using the first equations of motion,

                                 v = u + at

                                 v = 45 + (-7) (3)                      ∵  (-7 ) car is decelerating

                                  v =24 m/s

The final kinetic energy of the car

                                 k.e = 0.5 x 2000 x 24

                                        = 576000 J

The difference in the kinetic energy,

                           E = K.E - k.e

                               = 2025000 J - 576000 J

                               = 1449000 joules

Hence, the change in the Kinetic Energy of the car, E = 1449000 joules

3 0
3 years ago
Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo
Sonbull [250]

Answer:

2.267

Explanation:

Drag force is given by

F=\dfrac{1}{2}\rho Av^2C

C = Drag coefficient is constant

A = Area is constant

v_1 = Velocity of the passenger jet = 1200 km/h = \dfrac{1200}{3.6}\ \text{m/s}

v_2 = Velocity of the prop plane = \dfrac{1}{4}v_1

\rho_1 = Density of the air where the jet was flying = 0.38\ \text{kg/m}^3

\rho_2 = Density of the air where the prop plane was flying = 0.67\ \text{kg/m}^3

F\propto \rho v^2

\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267

The ratio of the drag forces is 2.267.

5 0
3 years ago
A constant force of 4.0 N acts on a variety of objects. The greatest acceleration will occur in the object that has
zaharov [31]
In the object that has, the least inertia.
8 0
3 years ago
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