All categories

3 years ago

A 3000 kg deck is cantilevered over a beam as shown. It is attached to the house at the other end. Find the force of the beam.

Physics

1 answer:

Afina-wow [57]3 years ago

4 0

The beam is acting as the pivot and the weight of the deck acts through its center its gravity, at the midpoint of the deck.
Center of gravity at: (16 + 4)/2 = 10 ft

Converting 4 ft to meters: 1.22 m
Converting 10 ft to meters: 3.05

F x 1.22 = 3.05 x 3000 x 9.81
F = 73,600 N

You might be interested in

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

Tresset [83]

Answer:231.16 N/C

Explanation:

Given

Electric Flux $=147 N-m^2/C$

Area(A) $=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6= $58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C

4 0

3 years ago

the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

dem82 [27]

Answer:

Time = 5 seconds

Distance = 50 meters

Explanation:

Constantly Accelerated Motion

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

$v_f=v_o+a.t$

And the distance s is

$\displaystyle s=v_o.t+\frac{gt^2}{2}$

Given the object starts from rest, vo=0 and vf=20 m/s at $a=4\ m/s^2$ . We compute t

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}$

$\boxed{t=5\ sec}$

Now we compute s

$\displaystyle s=0+\frac{4\times 5^2}{2}$

$\boxed{s=50\ m}$

5 0

2 years ago

Correctly round the following number to the ten thousandths place 3.68274

Sergeu [11.5K]

The answer to this question is 3.69

5 0

2 years ago

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

babunello [35]

A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.

Read more on Brainly.com - brainly.com/question/18743384#readmore

8 0

2 years ago

When using the scientist method, which step comes last?

iogann1982 [59]

Here, I hope this helps.

6 0

2 years ago