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suter [353]
3 years ago
14

State and explain in terms of the forces acting how the acceleration of the object varies during its fall?

Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer: State and explain, in terms of the forces acting,how the acceleration of the object varies during its fall. As the object is released two forces act on it. Its weight and and air resistance. At this point its weight is much greater due to the force of gravity on the object so it accelerates downward.

https://www.mytutor.co.uk › Physics

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
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