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HACTEHA [7]
1 year ago
5

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Physics
1 answer:
Elanso [62]1 year ago
6 0

The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

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in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?​
Pavlova-9 [17]

Answer:

There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.

Explanation:

3 0
3 years ago
When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of
Fittoniya [83]

Answer:

  75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

  KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

  KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

__

The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0
2 years ago
Suppose there is a sample of xenon in a sealed rectangular container. The gas exerts a total force of 6.05 N perpendicular to on
shtirl [24]

Answer:

<h2> 0.147136N/m²</h2>

Explanation:

Pressure is defines as force exerts by a body per its unit area.

Pressure = Force/Area

Given the total force exerted by the gas = 6.05N

Area of the rectangular container = 0.121 m * 0.201 m = 0.024321m²

Pressure of the sample = 6.05/0.02432

Pressure of the sample = 0.147136N/m²

6 0
3 years ago
There are great similarities between electric and gravitational fields. A room can be electrically shielded so that there are no
ella [17]

Answer:

Can a room be gravitationally shielded? No, it can't.

Explanation:

the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.

7 0
3 years ago
A horizontal spring with spring constant 200N/m is compressed by 15cm and used to launch a 2kg box across a frictionless horizon
kobusy [5.1K]

Answer:

Explanation:

Given that,

Spring constant k=200N/m

Compression x = 15cm = 0.15m

Attached mass m =2kg

Coefficient of kinetic friction uk= 0.2

The energy in the spring is given as

U =½kx²

U = ½ × 200 × 0.15²

U = 2.25J

Force in the spring is given by Hooke's law

F = ke

F = 200×0.15

F = 30N

The weight of body which is equal to the normal is give as

W = mg

W = 2 × 9.81

W = 19.62N

W = N = 19.62 Newton's 2nd Law

From law of friction,

Fr = uk•N

Fr = 0.2 × 19.62

Fr = 3.924

Using newton second law again

Fnet = F - Fr

Fnet = 30 - 3.924

Fnet = 26.076

Work done by net force is given as

W = Fnet × d

W = 26.076d

Then, the work done by this net force is equal to the energy in the spring

W = U

26.076d = 2.25

d = 2.25/26.076

d = 0.0863m

Which is 8.63cm

So the box will slide 8.63cm before stopping

6 0
3 years ago
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