Ask question

Ask question

All categories

Nutka1998 [239]

2 years ago

5

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

able is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam is 8.662x10^1 N.
What is the magnitude of the force that the beam exerts on the hi.nge?

1 answer:

irinina [24]2 years ago

7 0

288.51 N is the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The horizontal component of force + the vertical component of force

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.

Learn more about components of forces here brainly.com/question/26446720

#SPJ1

You might be interested in

A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the

tatiyna

Answer:

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁)................... Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t................... Equation 2

make t the subject of the equation

t = Q/P................. Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.

6 0

3 years ago

Which statement about gravity is always True? A. The force of gravity between two objects is greater when the objects have less

garri49 [273]

The answer is c because the farther apart they are the greater there gravity is

4 0

3 years ago

Read 2 more answers

In order to design an experiment, you need a ____ about the scientific question you are trying to answer.

goldenfox [79]

In order to design an experiment, you need a hypothesis about the scientific question you are trying to answer.

6 0

3 years ago

Read 2 more answers

A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe

Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

= (1410 - 207)/(80.5/60)

= 60(1410 - 207)/80.5

= 60(1203)80.5

= 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

= 207 × 1.342 + 1/2 × 896.65 × 1.342²

= 277.725 + 807.417

= 1085.14 revolutions ≅ 1085 revolutions

5 0

3 years ago

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

6 0

3 years ago

Read 2 more answers