If you hang a car of mass 1560 kg from a steel beam, the beam bends with an angle of 0.055°. What is the tension in this beam in newtons?

Answers

Answer 1

Answer:

Answer:

The tension in the steel beam is 14.72 Newtons.

Explanation:

To calculate the tension in the steel beam when a car is hanging from it, you can use the principles of static equilibrium. In this situation, the gravitational force acting on the car must be balanced by the tension in the steel beam.

First, let's calculate the gravitational force acting on the car:

F_gravity = mass × gravity

Where:

Mass (m) = 1560 kg

Gravity (g) ≈ 9.81 m/s² (standard acceleration due to gravity)

F_gravity = 1560 kg × 9.81 m/s² ≈ 15306 N

Now, this gravitational force is balanced by the tension in the steel beam. Since the beam bends with an angle of 0.055°, we need to consider the vertical component of the tension force.

The vertical component of the tension (T_vertical) can be calculated using trigonometry (considering the angle θ):

T_vertical = T × sin(θ)

Where:

T_vertical is the vertical component of tension.

T is the tension in the beam.

θ is the angle in radians.

We need to convert the angle from degrees to radians:

θ = 0.055° × (π/180) ≈ 0.000959 radians

Now, we can calculate T_vertical:

T_vertical = 15306 N × sin(0.000959) ≈ 14.72 N

So, the tension in the steel beam is 14.72 Newtons.

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A 1300-turn coil of wire that is 2.10 cm in diameter is in a magnetic field that drops from 0.130 T to 0 T in 12.0 ms. The axis of the coil is parallel to the field.Question: What is the emf of the coil? (in V)

Answers

Answer:

4.875 V

Explanation:

N = 1300

diameter = 2.10 cm

radius = half of diameter = 1.05 cm

B1 = 0.130 T

B2 = 0 T

t = 12 ms

According to the law of electromagnetic induction,

$e = - N(d\phi )/(dt)$

Where, Ф be the magnetic flux linked with the coil

$e = - NA (dB )/(dt)$

$e = -1300*3.14*{1.05* 1.05* 10^(-4)*(0-0.130)/(12*10^(-3))=$

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A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

Answers

Answer:

19.3 m/s

Explanation:

Take down to be positive. Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

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A physicist is creating a computational model of a falling person before and after opening a parachute. What boundary conditions would be important here?the air resistance encountered as the person falls

the speed at which the person falls

the change in kinetic and potential energy

the location where potential energy is zero

Answers

Answer:

the location where potential energy is zero

Explanation:

Answer:

Air resistance

Explanation:

Air resistance encountered as the person falls

Two blocks of masses 1 = 700 and 2 = 1100 are connected by a cord of negligible mass and hung over a diskshaped pulley, as shown in the figure. The pulley has a mass of = 1.50 and a radius of = 14 , and rotates about a lightweight axle through its center. The axle itself is hung from the ceiling by two like cords of negligible mass and is held horizontally. The system is released from rest. a) Draw a free-body diagram for each of the blocks and the pulley separately. b) Find the magnitude of the acceleration of the blocks. c) Find the magnitude of the angular acceleration of the pulley. d) Find the magnitude of tensions in the cords, 1, 2, and 3. (See the figure.)

Answers

Answer:

b) 16 cm

Magnification, m = v/u

3 = v/u

⇒ v = 3u

Lens formula : 1/v – 1/u = 1/f

1/3u = 1/u = 1/12

-2/3u = 1/12

⇒ u = -8 cm

V = 3 × (-8) = -24

Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm

Explanation:

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

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Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

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